# T-product Tensors—Part I: Inequalities

Shih Yu Chang Shih Yu Chang is with the Department of Applied Data Science,
San Jose State University, San Jose, CA, U. S. A. (e-mail:
shihyu.chang@sjsu.edu).

###### Abstract

The T-product operation between two three-order tensors was invented around
2011 and it arises from many applications, such as signal processing, image
feature extraction, machine learning, computer vision, and the multi-view
clustering problem. Although there are many pioneer works about T-product
tensors, there are no works dedicated to inequalities associated with
T-product tensors. In this work, we first attempt to build inequalities at the
following aspects: (1) trace function nondecreasing/convexity; (2) Golden-
Thompson inequality for T-product tensors; (3) Jensen’s T-product inequality;
(4) Klein’s T-product inequality. All these inequalities are related to
generalize celebrated Lieb’s concavity theorem from matrices to T-product
tensors. This new version of Lieb’s concavity theorem under T-product tensor
will be used to determine the tail bound for the maximum eigenvalue induced by
independent sums of random Hermitian T-product, which is the key tool to
derive various new tail bounds for random T-product tensors. Besides, Qi et.
al [1] introduces a new concept, named _eigentuple_ , about T-product tensors
and they apply this concept to study nonnegative (positive) definite
properties of T-product tensors. The final main contribution of this work is
to develop the Courant-Fischer Theorem with respect to eigentuples, and this
theorem helps us to understand the relationship between the minimum eigentuple
and the maximum eigentuple. The main content of this paper is Part I of a
serious task about T-product tensors. The Part II of this work will utilize
these new inequalities and Courant-Fischer Theorem under T-product tensors to
derive tail bounds of the extreme eigenvalue and the maximum eigentuple for
sums of random T-product tensors, e.g., T-product tensor Chernoff and
T-product tensor Bernstein bounds.

Corresponding Address:

Shih Yu Chang

Department of Applied Data Science

San Jose State University

San Jose, CA 95130, U. S. A.

e-mail<EMAIL_ADDRESS>
Index terms— Hermitian T-product tensors, eigentuples, trace function, Lieb’s
concavity for T-product tensors, Courant-Fischer theorem for T-product
tensors.

##  1 Introduction

###  1.1 T-product Tensors

The T-product operation between two three order tensors was introduced by
Kilmer and her collaborators in [2, 3]. It has been shown as a powerful tool
in many fields: signal processing [4, 5], machine learning [6], computer
vision [7, 8], low-rank tensor approximation [9, 10, 11] etc. Due to wide
applications of T-product, T-SVD and tubal ranks, Qi et. al [1]. extend
eigentuple concept first defined from [12] to study properties for symmetry of
T-product tensors and positive (nonnegative) semidefiniteness of T-product
tensors by defining a T-quadratic form, whose variable is an $m\times p$
matrix, and whose value is a $p$-dimensional vector. They further show that a
T-quadratic form is positive semidefinite (definite) if and only if the
smallest eigentuple of the corresponding T-symmetric tensor is nonnegative
(positive). However, none of these works discussed further issues about
inequalities associated with T-product tensors. The first part of this work
about T-product tensors is to develop several new T-product tensors
inequalities, which are the main topics discussed by this paper. In the matrix
setting, there are many useful applications about these similar inequalities
under the traditional matrix product, e.g., quantum information processing
[13]. The second part of this work is to apply these new inequalities about
T-product tensors to tail bounds estimation of the maximum eigenvalue and the
maximum eigentuple for sums of random T-product tensors, e.g., Chernoff and
Bernstein bounds. We will introduce these new inequalities about T-product
tensors obtained at this Part I work at the next subsection.

###  1.2 New Inequalities about T-product Tensors

In this work, we define trace, denoted by $\mathrm{Tr}$, as the summation of
f-diagonal entries of a given symmetric T-product tensor
$\mathcal{C}\in\mathbb{C}^{m\times m\times p}$ and study properties of trace
for T-product tensors. Our first main inequality about trace is following
theorem:

######  Theorem 1.1 (Monotonicity and Convexity of T-product Trace Function)

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function with non-
decreasing / convex / strictly convex properties, then so is the mapping
$\mathcal{C}\rightarrow\mathrm{Tr}\left(f(\mathcal{C})\right)$.

From trace definition, we will prove Golden–Thompson inequality for two
Hermitian T-product tensors which will be utilized to prove T-product tensors
martingale inequalities. A Hermitian T-product tensor is a tensor equal to its
Hermiitan transpose, which is defined by Eq. (19).

######  Theorem 1.2 (Golden-Thompson inequality for T-product Tensors)

Given two Hermitian T-product tensors
$\mathcal{C},\mathcal{D}\in\mathbb{C}^{m\times m\times p}$, we have

$\displaystyle\mathrm{Tr}\left(\exp(\mathcal{C}+\mathcal{D})\right)\leq\mathrm{Tr}\left(\exp\left(\mathcal{C}\right)\star\exp\left(\mathcal{D}\right)\right),$
(1)

where $\star$ is the product operation between two T-product tensors defined
by Eq. (27).

The next inequality we will show is Jensen’s operator inequality (positive
semidefinite relation between two T-product tensors). A tensor with T-positive
definite (or T-positive semi-definite) will be abbreviated as TPD (or TPSD),
see Section 2.3 for its definition. If we have a TPSD relation between two
T-product tensors $\mathcal{C}$ and $\mathcal{D}$ represented as
$\mathcal{C}\preceq\mathcal{D}$, then the difference T-product tensor
$\left(\mathcal{D}-\mathcal{C}\right)$ is a T-positive semi-definite tensor.
Let $\mathcal{I}_{mmp}\in\mathbb{C}^{m\times m\times p}$ be the identity
tensor defined by Eq. (21).

######  Theorem 1.3 (Jensen’s T-product Inequality)

For a continuous _T-product tensor convex_ function $f$ defined on an interval
$\mathrm{I}$. The definition of _T-product tensor convex_ is given by Eq.
(71). we have the following TPSD relation for each natural number $n$:

$\displaystyle
f\left(\sum\limits_{i=1}^{n}\mathcal{C}_{i}^{\mathrm{H}}\star\mathcal{X}_{i}\star\mathcal{C}_{i}\right)\preceq\sum\limits_{i=1}^{n}\mathcal{C}_{i}^{\mathrm{H}}\star
f\left(\mathcal{X}_{i}\right)\star\mathcal{C}_{i},$ (2)

where $\mathcal{X}_{i}\in\mathbb{C}^{m\times m\times p}$ are bounded,
Hermitian T-product tensors with all eigenvalues in the interval $\mathrm{I}$
and tensors $\mathcal{C}_{i}$ satisfying
$\sum\limits_{i=1}^{n}\mathcal{C}_{i}^{\mathrm{H}}\star\mathcal{C}_{i}=\mathcal{I}_{mmp}$.

The immediate application of Theorem 1.1 is to prove Klein’s inequality for
T-product tensor.

######  Theorem 1.4 (Klein’s T-product Inequality)

For all $\mathcal{C},\mathcal{D}$ Hermitian T-product tensors and a
differentiable convex function $f:\mathbb{R}\rightarrow\mathbb{R}$ or for all
$\mathcal{C},\mathcal{D}$ Hermitian T-product tensors and a differentiable
convex function $f:(0,\infty)\rightarrow\mathbb{R}$, we have

$\displaystyle\mathrm{Tr}\left(f(\mathcal{C})-f(\mathcal{D})-(\mathcal{C}-\mathcal{D})\star
f^{\prime}(\mathcal{D})\right)\geq 0.$ (3)

In both situations, if $f$ is strictly convex, equality holds if and only if
$\mathcal{C}=\mathcal{D}$.

Previous theorems will help us to establish the following main theorem of this
paper about Lieb’s concavity for T-product tensors since tail bounds for sums
of random T-product tensors will be derived based on such concavity relation.

######  Theorem 1.5 (Lieb’s concavity theorem for T-product tensors)

Let $\mathcal{H}$ be a Hermitian T-product tensor. Following map

$\displaystyle\mathcal{A}\rightarrow\mathrm{Tr}e^{\mathcal{H}+\log\mathcal{A}}$
(4)

is concave on the positive-definite cone.

We are ready to present the theorem for the tail bound of the maximum
eigenvalue induced by independent sums of random Hermitian T-product tensors
and this theorem will play a key role to establish various new tail bounds of
the maximum eigenvalue generated by independent sums of random T-product
tensors.

######  Theorem 1.6 (Master Tail Bound for Independent Sum of Random T-product
Tensors for Eigenvalue)

Given a finite sequence of independent Hermitian T-product tensors
$\\{\mathcal{X}_{i}\\}$, we have

$\displaystyle\mathrm{Pr}\left(\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\theta\right)$
$\displaystyle\leq$
$\displaystyle\inf\limits_{t>0}\Big{\\{}e^{-t\theta}\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n}\log\mathbb{E}e^{t\mathcal{X}_{i}}\right)\Big{\\}}.$
(5)

Similarly, we can generalize master tail bound for independent sum of random
Hermitian T-product tensors with respect to eigenvalue from Theorem 1.6 to
eigentuple version by the following theorem 1.7. We begin with $\bigodot$
operation defined in Proposition 2.1 from work [1].

Let $\mathbf{a}=(a_{1},a_{2},\cdots,a_{p})^{\mathrm{T}}\in\mathbb{C}^{p}$,
then operator circ to the vector $\mathbf{a}$ can be defined as

$\displaystyle\mbox{circ}(\mathbf{a})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\left(\begin{array}[]{ccccc}a_{1}&a_{p}&a_{p-1}&\cdots&a_{2}\\\
a_{2}&a_{1}&a_{p}&\cdots&a_{3}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
a_{p}&a_{p-1}&a_{p-2}&\cdots&a_{1}\\\ \end{array}\right),$ (10)

and
$\mbox{circ}^{-1}(\mbox{circ}(\mathbf{a}))\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathbf{a}$. Suppose that $\mathbf{a},\mathbf{b}\in\mathbb{C}^{p}$,
we define

$\displaystyle\mathbf{a}\bigodot\mathbf{b}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mbox{circ}(\mathbf{a})\cdot\mathbf{b},$ (11)

where $\cdot$ is the standard matrix and vector multiplication. Then, we are
ready to present the following theorem.

######  Theorem 1.7 (Master Tail Bound for Independent Sum of Random T-product
Tensors for Eigentuple)

Given a finite sequence of independent random Hermitian T-product tensors
$\\{\mathcal{X}_{i}\\}$ such that $\mathcal{X}_{i}\in\mathbb{C}^{m\times
m\times p}$, if $\sum\limits_{i=1}^{n}t\mathcal{X}_{i}$ satisfies Eq. (122),
we have

$\displaystyle\mathrm{Pr}\left(\mathbf{d}_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\mathbf{b}\right)$
$\displaystyle\leq$ $\displaystyle\inf\limits_{t>0}\min\limits_{1\leq j\leq
p}\left\\{\frac{\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n}\log\mathbb{E}e^{t\mathcal{X}_{i}}\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\},$
(12)

where $e_{\bigodot}^{t\mathbf{b}}\in\mathbb{C}^{p}$ is the exponential for the
vector $t\mathbf{b}$ with respect to $\bigodot$ operation.

The last important theorem is the Courant-Fischer theorem for T-product
tensors. This theorem will be used to figure out the relationship between the
maximum eigentuple and the minimum eigentule of a T-product tensor.

######  Theorem 1.8 (Courant-Fischer Theorem under T-product)

Let $\mathcal{A}\in\mathbb{C}^{m\times m\times p}$ be a Hermitian T-product
tensor with eigentuples
$\mathbf{d}_{1}\geq\mathbf{d}_{2}\geq\cdots\geq\mathbf{d}_{n}$. Let
$\\{\mathbf{U}_{j}^{[l]}\\}\in\mathbb{C}^{m\times p}$ be orthnomal matrices
for $1\leq j\leq m$ and $0\leq l\leq p-1$, $S_{k}$ be the space spanned by
$\\{\mathbf{U}_{j}^{[l]}\\}$ for $1\leq j\leq k$ and $0\leq l\leq p-1$, and
$T_{k}$ be the space spanned by $\\{\mathbf{U}_{j}^{[l]}\\}$ for $k\leq j\leq
m$ and $0\leq l\leq p-1$. Then, we have

$\displaystyle\mathbf{d}_{k}$ $\displaystyle=$
$\displaystyle\max\limits_{\begin{subarray}{c}S_{k}\subseteq\mathbb{C}^{m\times
p}\\\ \dim(S_{k})=k\times p\end{subarray}}\min\limits_{\mathbf{X}\in
S_{k}}\left(\mathbf{X}^{\mathrm{H}}\star\mathcal{A}\star\mathbf{X}\right)\bigg{/}_{\bigodot}\left(\mathbf{X}^{\mathrm{H}}\star\mathbf{X}\right)$
(13) $\displaystyle=$
$\displaystyle\min\limits_{\begin{subarray}{c}T_{k}\subseteq\mathbb{C}^{m\times
p}\\\ \dim(T_{k})=(m-k+1)\times p\end{subarray}}\max\limits_{\mathbf{X}\in
T_{k}}\left(\mathbf{X}^{\mathrm{H}}\star\mathcal{A}\star\mathbf{X}\right)\bigg{/}_{\bigodot}\left(\mathbf{X}^{\mathrm{H}}\star\mathbf{X}\right),$

where $\bigg{/}_{\bigodot}$ is the division (inverse operation) under
$\bigodot$.

All these inequalities and maximum/minimum eigentuples relation about
T-product tensors will be utilized to derive a serious of new tail bounds for
the extreme eigenvalue and eigentuple for sums of random T-product tensors.
These new inequalities different from author Chang’s previous works about
bounds for sums of random tensors based on Einstein product [14, 15].

###  1.3 Paper Organization

The rest of this paper is organized as follows. In Section 2, basic notions of
T-product tensors are introduced. Lieb’s concavity theorem under T-product
will be studied in Section 3. General tail bounds for random T-product tensors
are provided in Section 4. Courant-Fischer Theorem under T-product is given in
Section 5. Finally, conclusion will be drawn in Section 6.

Nomenclature: The sets of complex and real numbers are denoted by $\mathbb{C}$
and $\mathbb{R}$, respectively. The symbol $\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}$ denotes mathematical definition.

##  2 T-product Tensors

In this section, we will review T-product operations briefly and discuss
related properties in Sec. 2.1. The T-SVD decomposition of T-product tensors
and T-Symmetric tensors will be presented in Sec. 2.2

###  2.1 What are T-product Tensors

For a third order tensor $\mathcal{C}\in\mathbb{C}^{m\times n\times p}$, we
define bcirc operation to the tensor $\mathcal{C}$ as:

$\displaystyle\mbox{bcirc}(\mathcal{C})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\left(\begin{array}[]{ccccc}\mathbf{C}^{(1)}&\mathbf{C}^{(p)}&\mathbf{C}^{(p-1)}&\cdots&\mathbf{C}^{(2)}\\\
\mathbf{C}^{(2)}&\mathbf{C}^{(1)}&\mathbf{C}^{(p)}&\cdots&\mathbf{C}^{(3)}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\mathbf{C}^{(p)}&\mathbf{C}^{(p-1)}&\mathbf{C}^{(p-2)}&\cdots&\mathbf{C}^{(1)}\\\
\end{array}\right),$ (18)

where $\mathbf{C}^{(1)},\cdots,\mathbf{C}^{(p)}\in\mathbb{C}^{m\times n}$ are
frontal slices of tensor $\mathcal{C}$. The inverse operation of bcirc is
denoted as $\mbox{bcirc}^{-1}$ with relation
$\mbox{bcirc}^{-1}(\mbox{bcirc}(\mathcal{C}))\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathcal{C}$.

For a third order tensor $\mathcal{C}\in\mathbb{C}^{m\times m\times p}$, we
define Hermitian transpose of $\mathcal{C}$, denoted by
$\mathcal{C}^{\mathrm{H}}$ , as

$\displaystyle\mathcal{C}^{\mathrm{H}}=\mbox{bcirc}^{-1}((\mbox{bcirc}(\mathcal{C}))^{\mathrm{H}}).$
(19)

And a tensor $\mathcal{D}\in\mathbb{C}^{m\times m\times p}$ is called a
Hermitian T-product tensor if $\mathcal{D}^{\mathrm{H}}=\mathcal{D}$.
Similarly, we also define ranspose of $\mathcal{C}$, denoted by
$\mathcal{C}^{\mathrm{T}}$ , as

$\displaystyle\mathcal{C}^{\mathrm{T}}=\mbox{bcirc}^{-1}((\mbox{bcirc}(\mathcal{C}))^{\mathrm{T}}).$
(20)

And a tensor $\mathcal{D}\in\mathbb{C}^{m\times m\times p}$ is called a
Symmetric T-product tensor if $\mathcal{D}^{\mathrm{T}}=\mathcal{D}$.

The identity tensor $\mathcal{I}_{mmp}\in\mathbb{C}^{m\times m\times p}$ can
be defined as:

$\displaystyle\mathcal{I}_{mmp}=\mbox{bcirc}^{-1}(\mathbf{I}_{mp}),$ (21)

where $\mathbf{I}_{mp}$ is the identity matrix in $\mathbb{R}^{mp\times mp}$.
A zero tensor, denoted as $\mathcal{O}_{mnp}\in\mathbb{C}^{m\times n\times
p}$, is a tensor that all elements inside the tensor as $0$.

In order to define the T-product operation, we need to define another kind of
operation over a third order tensor. For a third order tensor
$\mathcal{C}\in\mathbb{C}^{m\times n\times p}$, we define unfold operation to
the tensor $\mathcal{C}$ as:

$\displaystyle\mbox{unfold}(\mathcal{C})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\left(\begin{array}[]{c}\mathbf{C}^{(1)}\\\ \mathbf{C}^{(2)}\\\
\vdots\\\ \mathbf{C}^{(p)}\\\ \end{array}\right),$ (26)

where $\mbox{unfold}(\mathcal{C})\in\mathbb{C}^{mp\times n}$, and the inverse
operation of unfold is fold with the relation
$\mbox{fold}(\mbox{unfold}(\mathcal{C}))\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathcal{C}$. Given $\mathcal{C}\in\mathbb{C}^{m\times n\times p}$
and $\mathcal{D}\in\mathbb{C}^{n\times k\times p}$, we define the T-product
between $\mathcal{C}$ and $\mathcal{D}$ as

$\displaystyle\mathcal{C}\star\mathcal{D}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mbox{fold}(\mbox{bcirc}(\mathcal{D})\mbox{unfold}(\mathcal{D})),$
(27)

where $\mathcal{C}\star\mathcal{D}\in\mathbb{C}^{m\times k\times p}$.

###### Definition 1

Let $\mathcal{S}=(s_{ijk})\in\mathbb{C}^{m\times n\times p}$ be a f-diagonal
tensor, i.e., each frontal slice of tensor $\mathcal{S}$ is a diagonal matrix.
Let $\mathbf{s}_{i}=(s_{ii1},s_{ii2},\cdots,s_{iip})^{\mathrm{T}}$ be the
$ii-$th tube of $\mathcal{S}$ for $1\leq i\leq\min\\{m,n\\}$. The f-diagonal
tensor $\mathcal{S}$ is in its _standard form_ if
$\mathbf{s}_{1}\geq\mathbf{s}_{2}\geq\cdots\geq\mathbf{s}_{\min\\{m,n\\}}$,
where $\geq$ is the elementwise comparison between two vectors.

We define the T-product tensor _trace_ for a tensor
$\mathcal{C}=(c_{ijk})\in\mathbb{C}^{m\times m\times p}$, denoted by
$\mathrm{Tr}(\mathcal{C})$, as following

$\displaystyle\mathrm{Tr}(\mathcal{C})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\sum\limits_{i=1}^{m}\sum\limits_{k=1}^{p}c_{iik},$ (28)

which is the summation of all entries in f-diagonal components. Then, we have
following lemma about trace properties.

###### Lemma 1

For any tensors $\mathcal{C},\mathcal{D}\in\mathbb{C}^{m\times m\times p}$, we
have

$\displaystyle\mathrm{Tr}(c\mathcal{C}+d\mathcal{D})=c\mathrm{Tr}(\mathcal{C})+d\mathrm{Tr}(\mathcal{D}),$
(29)

where $c,d$ are two contants. And, the transpose operation will keep the same
trace value, i.e.,

$\displaystyle\mathrm{Tr}(\mathcal{C})=\mathrm{Tr}(\mathcal{C}^{\mathrm{T}}).$
(30)

Finally, we have

$\displaystyle\mathrm{Tr}(\mathcal{C}\star\mathcal{D})=\mathrm{Tr}(\mathcal{D}\star\mathcal{C}).$
(31)

Proof: Eqs. (29) and (30) are true from trace definiton directly.

From Eq. (27), the $i$-th frontal slice matrix of
$\mathcal{D}\star\mathcal{C}$ is

$\displaystyle\mathbf{D}^{(i)}\mathbf{C}^{(1)}+\mathbf{D}^{(i-1)}\mathbf{C}^{(2)}+\cdots+\mathbf{D}^{(1)}\mathbf{C}^{(i)}+\mathbf{D}^{(m)}\mathbf{C}^{(i+1)}+\cdots+\mathbf{D}^{(i+1)}\mathbf{C}^{(m)},$
(32)

similarly, the $i$-th frontal slice matrix of $\mathcal{C}\star\mathcal{D}$ is

$\displaystyle\mathbf{C}^{(i)}\mathbf{D}^{(1)}+\mathbf{C}^{(i-1)}\mathbf{D}^{(2)}+\cdots+\mathbf{C}^{(1)}\mathbf{D}^{(i)}+\mathbf{C}^{(m)}\mathbf{D}^{(i+1)}+\cdots+\mathbf{C}^{(i+1)}\mathbf{D}^{(m)}.$
(33)

Because the matrix trace of Eq. (32) and the matrix trace of Eq. (33) are same
for each slice $i$ due to linearity and invariant under cyclic permutations of
matrix trace, we have Eq. (31) by summing over all frontal matrix slices.
$\hfill\Box$

Below, we will define the _determinant_ of a T-product tensor
$\mathcal{C}\in\mathbb{C}^{m\times m\times p}$ and its asscoiate properties.
The determinant of a $m\times m\times p$ tensor $\mathcal{C}$ is the
$m$-linear alternating form defined as

$\displaystyle\mathrm{det}:\left(\mathbf{V}_{1},\cdots,\mathbf{V}_{m}\right)\rightarrow\mathbb{C},$
(34)

where $\mathbf{V}_{i}\in\mathbb{C}^{m\times p}$ is the $i$-th lateral matrix
of the tensor $\mathcal{C}$. Moreover, we require that
$\mathrm{det}(\mathcal{I}_{mmp})=1$. Given two tensors
$\mathcal{C},\mathcal{D}\in\mathbb{C}^{m\times m\times p}$, the determinant of
$\mathcal{C}\star\mathcal{D}$ is
$\mathrm{det}(\mathcal{C}\star\mathcal{D})=\lambda\mathrm{det}(\mathcal{D})$
for some value $\lambda$. If we set $\mathcal{D}$ as $\mathcal{I}_{mmp}$, we
have

$\displaystyle\mathrm{det}(\mathcal{C}\star\mathcal{I}_{mmp})=\lambda\mathrm{det}(\mathcal{I}_{mmp})=\lambda=\mathrm{det}(\mathcal{C}).$
(35)

Then, we have

$\displaystyle\mathrm{det}(\mathcal{C}\star\mathcal{D})=\mathrm{det}(\mathcal{C})\mathrm{det}(\mathcal{D})$
(36)

###  2.2 T-SVD Decomposition

Given a tensor $\mathcal{C}\in\mathbb{C}^{m\times n\times p}$, Theorem 4.1 in
[2] proposed a T-singular value decomposition (T-SVD) for $\mathcal{C}$ as:

$\displaystyle\mathcal{C}=\mathcal{U}\star\mathcal{S}\star\mathcal{V}^{\mathrm{T}},$
(37)

where $\mathcal{U}\in\mathbb{C}^{m\times m\times p}$ and
$\mathcal{V}\in\mathbb{C}^{n\times n\times p}$ are orthogonal tensors, and
$\mathcal{S}\in\mathbb{C}^{m\times n\times p}$ is a f-diagonal tensor. We also
have $\mathcal{U}^{\mathrm{T}}\star\mathcal{U}=\mathcal{I}_{mmp}$ and
$\mathcal{V}^{\mathrm{T}}\star\mathcal{V}=\mathcal{I}_{nnp}$. We define
$\sigma(\mathcal{C})$ be the spectrum of $\mathcal{C}$, i.e., the set of
$s\in\mathbb{C}$, where $s$ are nonzero entries from tensor $\mathcal{S}$. We
use $\left\|\cdot\right\|$ for the spectral norm, which is the largest
singular value of a T-product tensor.

Given any integer $k$ and $\mathcal{B}\in\mathbb{C}^{m\times m\times p}$, we
define $\mathcal{B}^{k}$ as

$\displaystyle\mathcal{B}^{k}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\overbrace{\mathcal{B}\star\mathcal{B}\star\mathcal{B}\star\cdots\star\mathcal{B}}^{\mbox{$k$
terms of $\mathcal{B}$ under T-product}}$ (38)

where $\mathcal{B}^{k}\in\mathbb{C}^{m\times m\times p}$. Then, we have
following corollary from T-SVD in Eq. (37).

###### Corollary 1

Suppose $\mathcal{B}\in\mathbb{C}^{m\times m\times p}$ is a Hermitian
T-product tensor, and $\mathcal{S}^{-1}$ exists, where $\mathcal{S}$ is
f-diagonal tensor obtained from the T-SVD of the tensor $\mathcal{C}$. Then,
we have

$\displaystyle\mathcal{B}^{k}=\mathcal{U}\star\mathcal{S}^{k}\star\mathcal{U}^{\mathrm{T}}.$
(39)

Then, we can define the T-product tensor exponential function and the
T-product tensor logarithm function under T-product as below with tensor
power.

###### Definition 2

Given a tensor $\mathcal{X}\in\mathbb{C}^{m\times m\times p}$, the _tensor
exponential_ of the tensor $\mathcal{X}$ is defined as

$\displaystyle e^{\mathcal{X}}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\sum\limits_{k=0}^{\infty}\frac{\mathcal{X}^{k}}{k!},$ (40)

where $\mathcal{X}^{0}$ is defined as the identity tensor $\mathcal{I}_{mmp}$.
Given a tensor $\mathcal{Y}$, the tensor $\mathcal{X}$ is said to be a _tensor
logarithm_ of $\mathcal{Y}$ if $e^{\mathcal{X}}=\mathcal{Y}$.

From T-SVD in Eq. (37), we can express a Hermitian T-product tensor
$\mathcal{C}\in\mathbb{C}^{m\times m\times p}$ as

$\displaystyle\mathcal{C}=\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}s_{iik}\mathbf{U}_{i}^{[k]}\star\left(\mathbf{U}_{i}^{[k]}\right)^{\mathrm{T}},$
(41)

where $s_{iik}$ are _eigenvalues_ of the tensor $\mathcal{C}$, and
$\mathbf{U}_{i}^{[k]}\in\mathbb{C}^{m\times 1\times p}$ is the $i$-th lateral
slice (matrix) of the tensor $\mathcal{U}$ after $k$ cyclic permutations. The
matrix $\mathbf{U}_{i}^{[0]}$ is obtained from the $i$-th lateral slice
(matrix) of the tensor $\mathcal{U}$ with column vectors as
$\mathbf{u}_{i}^{(1)},\cdots,\mathbf{u}_{i}^{(p)}$, then we have

$\displaystyle\mathbf{U}_{i}^{[k]}=\left(\mathbf{u}_{i}^{(p+1-k)\bmod
p},\mathbf{u}_{i}^{(p+2-k)\bmod
p},\cdots,\mathbf{u}_{i}^{(p)},\mathbf{u}_{i}^{(1)},\cdots\mathbf{u}_{i}^{(p-k)}\right).$
(42)

Note that we have
$\left(\mathbf{U}_{i}^{[k]}\right)^{\mathrm{H}}\star\mathbf{U}_{i}^{[k]}=\mathcal{I}_{11p}$
and
$\left(\mathbf{U}_{i}^{[k]}\right)^{\mathrm{H}}\star\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}=\mathcal{O}_{11p}$
for $i\neq i^{\prime}$ or $k\neq k^{\prime}$. From Theorem 3.6 in [1], all
values of $s_{iik}$ are real and we define
$\lambda_{\max}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\max\limits_{\begin{subarray}{c}1\leq i\leq m\\\ 0\leq k\leq
p-1\end{subarray}}\\{s_{iik}\\}$, and
$\lambda_{\min}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\min\limits_{\begin{subarray}{c}1\leq i\leq m\\\ 0\leq k\leq
p-1\end{subarray}}\\{s_{iik}\\}$.

From Corollary 1 and Eq. (41), we can have following spectral mapping lemma.

###### Lemma 2

For any continous function $f:\mathbb{R}\rightarrow\mathbb{R}$ and any
Hermitian T-product tensor $\mathcal{C}$, we have

$\displaystyle
f(\mathcal{C})=\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}f(s_{iik})\mathbf{U}_{i}^{[k]}\star\left(\mathbf{U}_{i}^{[k]}\right)^{\mathrm{T}}.$
(43)

###  2.3 Positive Semidefinite T-product Tensors

Given a Hermitian T-product tensor $\mathcal{C}\in\mathbb{C}^{m\times m\times
p}$, and a tensor $\mathcal{X}\in\mathbb{C}^{m\times 1\times p}$ obtained from
treating the matrix $\mathbf{X}\in\mathbb{C}^{m\times p}$ as a tensor with
dimensions $\mathbb{R}^{m\times 1\times p}$. We define following quadratic
form with respect to the matrix $\mathbf{X}$ as

$\displaystyle F_{\mathcal{C}}(\mathbf{X})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathcal{X}^{\mathrm{T}}\star\mathcal{C}\star\mathcal{X},$ (44)

and we say that a tensor $\mathcal{C}$ is T-positive definite (TPD) (or
T-positive semi-definite (TPSD)) if $F_{\mathcal{C}}(\mathbf{X})>\mathbf{0}$
(or $F_{\mathcal{C}}(\mathbf{X})\geq\mathbf{0}$ ) for any
$\mathbf{X}\in\mathbb{C}^{m\times p}$, where $\mathbf{0}$ is a zero vector
with size $p$.

We now define eigentuples and eigenmatrices of a Hermitian T-product tensor
which will be used to characterize TPD or TPSD for a given tensor. For a
matrix $\mathbf{X}\in\mathbb{C}^{m\times
p}=(\mathbf{x}^{(1)},\cdots,\mathbf{x}^{(p)})$ , we define unfolding opeartion
with respect to the matrix $\mathbf{X}$ columns, denoted by
$\mbox{cunfold}(\mathbf{X})$, as

$\displaystyle\mbox{cunfold}(\mathbf{X})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\cdot\left(\begin{array}[]{c}\mathbf{x}^{(1)}\\\
\mathbf{x}^{(2)}\\\ \vdots\\\ \mathbf{x}^{(p)}\\\ \end{array}\right),$ (49)

where $\mbox{cunfold}(\mathbf{X})\in\mathbb{C}^{mp}$. Then, suppose that
$\mathcal{C}\in\mathbb{C}^{m\times m\times p}$ is a Hermitian T-product
tensor, we define $\mathcal{C}\star\mathbf{X}$ as

$\displaystyle\mathcal{C}\star\mathbf{X}=\mbox{fold}(\mbox{bcirc}(\mathcal{C})\mbox{cunfold}(\mathbf{X})),$
(50)

where $\mathcal{C}\star\mathbf{X}\in\mathbb{C}^{m\times p}$. We also define a
new product operation between a vector
$\mathbf{d}=(d_{1},d_{2},\cdots,d_{p})^{\mathrm{T}}$ and a matrix
$\mathbf{X}\in\mathbb{C}^{m\times p}$, denoted by $\circ$, as

$\displaystyle\mathbf{d}\circ\mathbf{X}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathbf{X}\cdot\left(\begin{array}[]{ccccc}d_{1}&d_{p}&d_{p-1}&\cdots&d_{2}\\\
d_{2}&d_{1}&d_{p}&\cdots&d_{3}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
d_{p}&d_{p-1}&d_{p-2}&\cdots&d_{1}\\\ \end{array}\right),$ (55)

where $\cdot$ is the standard matrix multiplication. Suppose that
$\mathbf{X}\in\mathbb{C}^{m\times p}$ and $\mathbf{X}\neq\mathbf{O}$, and
$\mathbf{d}\in\mathbb{C}^{p}$, if we have

$\displaystyle\mathcal{C}\star\mathbf{X}=\mathbf{d}\circ\mathbf{X},$ (56)

we call $\mathbf{d}$ as an eigentuple of $\mathcal{C}$, and $\mathbf{X}$ as an
eigenmatrix of $\mathcal{C}$ corresponding to the eigentuple $\mathbf{d}$.

From Theorem 4.1 in [1], a T-square tensor $\mathcal{C}\in\mathbb{C}^{m\times
m\times p}$ with eigentuples arranged as f-diagonal tensor $\mathcal{S}$
according to the standard form provided by Definition 1, i.e.,
$\mathbf{s}_{1}\geq\mathbf{s}_{2}\geq\cdots\geq\mathbf{s}_{m}$. Then
$\mathcal{C}$ is TPD (or TPSD) if and only if the smallest eigentuple
$\mathbf{s}_{m}>(\mbox{or $\geq$ })\mathbf{0}$. We use
$\left\|\mathcal{C}\right\|_{\mbox{\tiny{vec}}}$ to represent the spectral
norm of eigentuple of the tensor $\mathcal{C}$, which is defined as

$\displaystyle\left\|\mathcal{C}\right\|_{\mbox{\tiny{vec}}}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathbf{d}_{\max}\left(\sqrt{\mathcal{C}^{\mathrm{H}}\star\mathcal{C}}\right).$
(57)

###  2.4 T-product Tensors Analysis

We will begin with monotonicity and convexity discussions of the trace
function.

###### Lemma 3

For a given continous and non-decreasing function
$f:\mathbb{R}\rightarrow\mathbb{R}$, the associated trace function on a
Hermitian T-product tensor $\mathcal{C}$ is given by

$\displaystyle\mathcal{C}\rightarrow\mathrm{Tr}\left(f(\mathcal{C})\right).$
(58)

Then we have

$\displaystyle\mathcal{C}\succeq\mathcal{D}\Longrightarrow\mathrm{Tr}\left(f(\mathcal{C})\right)\geq\mathrm{Tr}\left(f(\mathcal{D})\right).$
(59)

Proof: We first assume that the function $f$ is differentiable, then the first
derivative of $f$ is greater or equal than zero (monotonicity). We further
define a trace function $g(t)\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathrm{Tr}\left(f\left(\mathcal{D}+t(\mathcal{C}-\mathcal{D})\right)\right)$.
Then, we have

$\displaystyle\mathrm{Tr}\left(f(\mathcal{C})\right)-\mathrm{Tr}\left(f(\mathcal{D})\right)=g(1)-g(0)=\int\limits_{0}^{1}g^{\prime}(t)dt=\int\limits_{0}^{1}\mathrm{Tr}\left(f^{\prime}(\mathcal{D}+t(\mathcal{C}-\mathcal{D}))\star(\mathcal{C}-\mathcal{D})\right)dt$
$\displaystyle=\int\limits_{0}^{1}\mathrm{Tr}\left((\mathcal{C}-\mathcal{D})^{1/2}\star
f^{\prime}(\mathcal{D}+t(\mathcal{C}-\mathcal{D}))\star(\mathcal{C}-\mathcal{D})^{1/2}\right)\geq
0,\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ $ (60)

where we apply Lemma 1 at the last equality, and the last inequality comes
from the nonnegative of $f^{\prime}$. By applying the standard continuity
argument, we can relax the requirement that $f$ is continuously differentiable
to the requirement that $f$ is continuous. $\hfill\Box$

The next lemma will be used to show the convexity of trace function on a
Hermitian T-product tensor $\mathcal{C}$.

###### Lemma 4

Let $\mathcal{C}\in\mathbb{C}^{m\times m\times p}$ be a Hermitian T-product
tensor, $f$ convex on $\mathbb{R}$, and $\mathbf{V}_{i}^{[k]}$ for $1\leq
i\leq m$ and $0\leq k\leq p-1$ be any orthnormal base of $\mathbb{C}^{m\times
p}$. Then, we have

$\displaystyle\mathrm{Tr}\left(f(\mathcal{C})\right)\geq\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}f\left(\left\langle\mathbf{V}_{i}^{[k]},\mathcal{C}\star\mathbf{V}_{i}^{[k]}\right\rangle\right),$
(61)

where
$\left\langle\mathbf{V}_{i}^{[k]},\mathcal{C}\star\mathbf{V}_{i}^{[k]}\right\rangle$
is the Frobenius inner product between two matrices $\mathbf{V}_{i}^{[k]}$ and
$\mathcal{C}\star\mathbf{V}_{i}^{[k]}$. There is an equality if each
$\mathbf{V}_{i}^{[k]}$ is an eigenmatrix of $\mathcal{C}$ and it’s the only
case if $f$ is strictly convex.

Proof: From spectral representation by Eq. (41), we have

$\displaystyle\mathrm{Tr}\left(f(\mathcal{C})\right)$ $\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}\left\langle\mathbf{V}_{i}^{[k]},\sum\limits_{i^{\prime}=1}^{m}\sum\limits_{k^{\prime}=0}^{p-1}f(s_{i^{\prime}i^{\prime}k^{\prime}})\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\star\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\right)^{\mathrm{T}}\right)\star\mathbf{V}_{i}^{[k]}\right\rangle$
(62) $\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}\sum\limits_{i^{\prime}=1}^{m}\sum\limits_{k^{\prime}=0}^{p-1}f(s_{i^{\prime}i^{\prime}k^{\prime}})\left\|\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\star\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\right)^{\mathrm{T}}\right)\star\mathbf{V}_{i}^{[k]}\right\|^{2}$
$\displaystyle\geq$
$\displaystyle\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}f\left(\sum\limits_{i^{\prime}=1}^{m}\sum\limits_{k^{\prime}=0}^{p-1}s_{i^{\prime}i^{\prime}k^{\prime}}\left\langle\mathbf{V}_{i}^{[k]},\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\star\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\right)^{\mathrm{T}}\right)\star\mathbf{V}_{i}^{[k]}\right\rangle\right)$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}f\left(\left\langle\mathbf{V}_{i}^{[k]},\mathcal{C}\star\mathbf{V}_{i}^{[k]}\right\rangle\right),$

where the only inequality comes from the convexity of the function $f$. Since
for each $i,k$, we have
$\sum\limits_{i^{\prime}=1}^{m}\sum\limits_{k^{\prime}=0}^{p-1}\left\|\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\star\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\right)^{\mathrm{T}}\right)\star\mathbf{V}_{i}^{[k]}\right\|^{2}=\left\|\mathbf{V}_{i}^{[k]}\right\|^{2}=1$.
Note that each $\mathbf{V}_{i}^{[k]}$ is an eigenmatrix of $\mathcal{C}$ if
and only if
$\left\|\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\star\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\right)^{\mathrm{T}}\right)\star\mathbf{V}_{i}^{[k]}\right\|^{2}=1$
for some $i^{\prime},k^{\prime}$, and is 0 otherwise, in which case the
inequality in Eq. (61) is eqaulity. When $f$ is strictly convex, equality in
Eq. (61) can be true only if for each $i,k$, we have
$\left\|\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\star\left(\mathbf{U}_{i^{\prime}}^{[k^{\prime}]}\right)^{\mathrm{T}}\right)\star\mathbf{V}_{i}^{[k]}\right\|^{2}=1$
for some $i^{\prime},k^{\prime}$, and is 0 otherwise. $\hfill\Box$

From Lemma 3 and Lemma 4, we have the following theorem about convexity and
monotonicity of a trace function. We recall 1.1.

See 1.1 Proof: Given two $\mathcal{C},\mathcal{D}\in\mathbb{C}^{m\times
m\times p}$ Hermitian T-product tensors, $f$ as a convex function, and
$\mathbf{V}_{i}^{[k]}$ for $1\leq i\leq m$ and $0\leq k\leq p-1$ be an
orthonormal basis of $\mathbb{C}^{m\times p}$ consisting of eigenmatrices of
$\frac{\mathcal{C}+\mathcal{D}}{2}$. Then, from Lemma 4, we have

$\displaystyle\mathrm{Tr}\left(f(\frac{\mathcal{C}+\mathcal{D}}{2})\right)$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}f\left(\left\langle\mathbf{V}_{i}^{[k]},\frac{\mathcal{C}+\mathcal{D}}{2}\star\mathbf{V}_{i}^{[k]}\right\rangle\right)$
(63) $\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}f\left(\frac{1}{2}\left\langle\mathbf{V}_{i}^{[k]},\mathcal{C}\star\mathbf{V}_{i}^{[k]}\right\rangle+\frac{1}{2}\left\langle\mathbf{V}_{i}^{[k]},\mathcal{D}\star\mathbf{V}_{i}^{[k]}\right\rangle\right)$
$\displaystyle\leq$
$\displaystyle\sum\limits_{i=1}^{m}\sum\limits_{k=0}^{p-1}\left(\frac{1}{2}f\left(\left\langle\mathbf{V}_{i}^{[k]},\mathcal{C}\star\mathbf{V}_{i}^{[k]}\right\rangle\right)+\frac{1}{2}f\left(\left\langle\mathbf{V}_{i}^{[k]},\mathcal{D}\star\mathbf{V}_{i}^{[k]}\right\rangle\right)\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\mathrm{Tr}\left(f\left(\mathcal{C}\right)\right)+\frac{1}{2}\mathrm{Tr}\left(f\left(\mathcal{D}\right)\right)$
(64)

where inequalities come from Lemma 4. This demonstrates that the map
$\mathcal{C}\rightarrow\mathrm{Tr}\left(f\left(\mathcal{C}\right)\right)$ is
midpoint convex.

For the strict convexity of $f$ and
$\mathrm{Tr}\left(f(\frac{\mathcal{C}+\mathcal{D}}{2})\right)=\frac{1}{2}\mathrm{Tr}\left(f\left(\mathcal{C}\right)\right)+\frac{1}{2}\mathrm{Tr}\left(f\left(\mathcal{D}\right)\right)$,
we have
$\left\langle\mathbf{V}_{i}^{[k]},\mathcal{C}\star\mathbf{V}_{i}^{[k]}\right\rangle=\left\langle\mathbf{V}_{i}^{[k]},\mathcal{D}\star\mathbf{V}_{i}^{[k]}\right\rangle$
for each $\mathbf{V}_{i}^{[k]}$. From Lemma 4, the equality will be true when
$\mathbf{V}_{i}^{[k]}$ are eigenmatrices for both tensors $\mathcal{C}$ and
$\mathcal{D}$. Then, we have

$\displaystyle\mathcal{C}\star\mathbf{V}_{i}^{[k]}=\langle\mathbf{V}_{i}^{[k]},\mathcal{C}\star\mathbf{V}_{i}^{[k]}\rangle\mathbf{V}_{i}^{[k]}=\langle\mathbf{V}_{i}^{[k]},\mathcal{D}\star\mathbf{V}_{i}^{[k]}\rangle\mathbf{V}_{i}^{[k]}=\mathcal{D}\star\mathbf{V}_{i}^{[k]},$
(65)

which indicates that $\mathcal{C}=\mathcal{D}$. An obvious continuity argument
now shows that if $f$ continuous as well as convex,
$\mathcal{C}\rightarrow\mathrm{Tr}\left(f(\mathcal{C})\right)$ is convex, and
strictly convex so if $f$ is strictly convex. Therefore, this Theorem is
proved from Lemma 3 and above arguments. $\hfill\Box$

From T-SVD, we have following relation for Hermitian T-product tensor:

$\displaystyle f(s)\leq g(s)\leavevmode\nobreak\ \leavevmode\nobreak\
\mbox{for $s\in[a,b]$}\Longrightarrow f(\mathcal{C})\preceq
g(\mathcal{C})\mbox{\leavevmode\nobreak\ \leavevmode\nobreak\ when the
eigenvalues of $\mathcal{C}$ lie in $[a,b]$.}$ (66)

Above Eq. (66) is named as transfer rule.

We have defined tensor exponential under Definition 2, and the exponential of
an Hermitian T-product tensor is always TPD due to the spectral mapping Lemma
2. From transfer rule Eq. (66), the tensor exponential satisfies following
relations for a Hermitian T-product tensor $\mathcal{C}\in\mathbb{C}^{m\times
m\times p}$ that we will use at later theory development:

$\displaystyle\mathcal{I}_{mmp}+\mathcal{C}\preceq\exp(\mathcal{C}),$ (67)

and

$\displaystyle\cosh(\mathcal{C})\preceq\exp(\mathcal{C}^{2}/2).$ (68)

From Theorem 1.1 and the monotonicity of the exponential function, we have

$\displaystyle\mathcal{C}\preceq\mathcal{D}\Longrightarrow\mathrm{Tr}\exp(\mathcal{C})\leq\mathrm{Tr}\exp(\mathcal{D})$
(69)

Below, we want to prove the monotonicity and the concavity of the logarithm
function. We will begin definitions about T-product tensor monotonicity and
convexity first and present several lemmas used to establish the monotonicity
and the concavity of the logarithm function. Given two Hermitian T-product
tensors $\mathcal{C},\mathcal{D}\in\mathbb{C}^{m\times m\times p}$, a function
$f:\mathbb{R}\rightarrow\mathbb{R}$ is said to be T-product tensor monotone if
the following relation holds:

$\displaystyle\mathcal{C}\preceq\mathcal{D}\Longrightarrow
f(\mathcal{C})\preceq f(\mathcal{D}).$ (70)

A function $f$ is said as a _T-product tensor convex_ function if we have:

$\displaystyle f(t\mathcal{C}+(1-t)\mathcal{D})\preceq
tf(\mathcal{C})+(1-t)f(\mathcal{D})(\mathcal{D}),$ (71)

where $0\leq t\leq 1$. Also, a function $f$ is said as a _T-product tensor
concave_ function if $-f$ is a _T-product tensor convex_ function. The
following derivation about the monotonicity and the concavity of the logarithm
function is extended from matrices according to works in [16] and [17] to
T-product tensors.

###### Lemma 5

For any $\mathcal{C},\mathcal{D}\in\mathbb{C}^{m\times m\times p}$, we have
$\sigma(\mathcal{C}\star\mathcal{D})=\sigma(\mathcal{D}\star\mathcal{C})$.

Proof: Since eigenvalues are roots of the characteristic polynomial, it is
enough to show that
$\mathrm{det}(\lambda\mathcal{I}_{mmp}-\mathcal{C}\star\mathcal{D})=\mathrm{det}(\lambda\mathcal{I}_{mmp}-\mathcal{D}\star\mathcal{C})$.
We first assume that $\mathcal{C}$ has inverse, then from Eq. (36), we have

$\displaystyle\mathrm{det}(\lambda\mathcal{I}_{mmp}-\mathcal{C}\star\mathcal{D})=\mathrm{det}(\mathcal{C}^{-1}\star(\lambda\mathcal{I}_{mmp}-\mathcal{C}\star\mathcal{D})\star\mathcal{C})=\mathrm{det}(\lambda\mathcal{I}_{mmp}-\mathcal{D}\star\mathcal{C}).$
(72)

This shows that
$\sigma(\mathcal{C}\star\mathcal{D})=\sigma(\mathcal{D}\star\mathcal{C})$.

If $\mathcal{C}$ is not invertible, we choose a sequence $\\{\epsilon_{n}\\}$
in $\mathbb{C}\backslash\sigma(\mathcal{C})$ with $\epsilon_{n}\rightarrow 0$,
with property that all new tensors
$\mathcal{C}_{n}\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathcal{C}-\epsilon_{n}\mathcal{I}_{mmp}$ are invertibale for each
$n$. Then,

$\displaystyle\mathrm{det}(\lambda\mathcal{I}_{mmp}-\mathcal{C}\star\mathcal{D})$
$\displaystyle=$
$\displaystyle\lim\limits_{n\rightarrow\infty}\mathrm{det}(\lambda\mathcal{I}_{mmp}-\mathcal{C}_{n}\star\mathcal{D})=\lim\limits_{n\rightarrow\infty}\mathrm{det}(\lambda\mathcal{I}_{mmp}-\mathcal{D}\star\mathcal{C}_{n})$
$\displaystyle=$
$\displaystyle\mathrm{det}(\lambda\mathcal{I}_{mmp}-\mathcal{D}\star\mathcal{C}).$
(73)

$\hfill\Box$

###### Lemma 6

For every tensor $\mathcal{C}\in\mathbb{C}^{m\times m\times p}$ and every
function $f$ on $\sigma(\mathcal{C}^{\mathrm{H}}\star\mathcal{C})$, we have

$\displaystyle\mathcal{C}\star f(\mathcal{C}^{\mathrm{H}}\star\mathcal{C})$
$\displaystyle=$ $\displaystyle
f(\mathcal{C}\star\mathcal{C}^{\mathrm{H}})\star\mathcal{C}.$ (74)

Proof: Since
$\sigma(\mathcal{C}^{\mathrm{H}}\star\mathcal{C})=\sigma(\mathcal{C}\star\mathcal{C}^{\mathrm{H}})$
from Lemma 5 and
$\mathcal{C}\star\left(\mathcal{C}^{\mathrm{H}}\star\mathcal{C}\right)^{n}=\left(\mathcal{C}\star\mathcal{C}^{\mathrm{H}}\right)^{n}\star\mathcal{C}$,
for $n\in\mathbb{N}$, this lemma is hold for $f$ is a polynomial. If the
function $f$ is an arbitary function on
$\sigma(\mathcal{C}^{\mathrm{H}}\star\mathcal{C})=[s_{1},\cdots s_{n}]$, we
define the Lagrance interpolation polynomial as

$\displaystyle p(x)\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\sum\limits_{i=1}^{n}f(s_{i})\prod\limits_{1\leq j\leq n,i\neq
j}\frac{x-s_{j}}{s_{i}-s_{j},}$ (75)

where we have $p(s_{i})=f(s_{i})$ for $1\leq i\leq n$. Then, we also have

$\displaystyle\mathcal{C}\star
f(\mathcal{C}^{\mathrm{H}}\star\mathcal{C})=\mathcal{C}\star
p(\mathcal{C}^{\mathrm{H}}\star\mathcal{C})=p(\mathcal{C}\star\mathcal{C}^{\mathrm{H}})\star\mathcal{C}=f(\mathcal{C}\star\mathcal{C}^{\mathrm{H}})\star\mathcal{C},$
(76)

and this Lemma is proved. $\hfill\Box$

Following Lemma is adopted from Corollary 12 from [17].

###### Lemma 7

We have following equivalent statements about a function
$f(x):(0,\infty)\rightarrow(0,\infty)$:

  1. 1.

$f(x)$ is T-product tensor monotone function;

  2. 2.

$x/f(x)$ is T-product tensor monotone function;

  3. 3.

$f(x)$ is T-product tensor concave function;

  4. 4.

$1/f(x)$ is T-product tensor convex function.

Proof: The proof is based in Corollary 12 from [17]. But those facts about
using Theorem 2.5.2 and Theorem 2.5.3 from [16] should be modified from
matrices settings to T-product tensors settings. With help from Lemma 6 and
transfer rules provided by Eq. (66), the proof about Theorem 2.5.2 and Theorem
2.5.3 from [16] for T-product tensors is straightforward by replacing matrix
multiplication opertions to T-product operations. $\hfill\Box$

We are ready to prove that the logarithmic function is T-product tensor
monotone and concave function on $(0,\infty)$.

###### Lemma 8

Given two TPD tensors $\mathcal{C},\mathcal{D}\in\mathbb{C}^{m\times m\times
p}$ with $\mathcal{O}\preceq\mathcal{C}\preceq\mathcal{D}$, we have

$\displaystyle\log(\mathcal{C})\preceq\log(\mathcal{D}),$ (77)

and

$\displaystyle
t\log(\mathcal{C})+(1-t)\log(\mathcal{D})\preceq\log(t\mathcal{C}+(1-t)\mathcal{D}).$
(78)

Proof: We define a function $g(x)=\frac{x}{\log(x+1)}$ on $(0,\infty)$. Since
$g(x)$ is the monotone function on $(0,\infty)$, Lemma 7 implies that
$\log(1+x)$ is T-product tensor monotone and concave on $(0,\infty)$. For each
$\epsilon>0$, $\log(\epsilon+x)=\log\epsilon+\log(1+x/\epsilon)$ is T-product
tensor monotone and concave on $(0,\infty)$. Let $\epsilon\rightarrow 0$, we
achive the desired result. $\hfill\Box$

In general, it is not practical to always working with Hermitian T-product
tensor, we will apply dilations techqnique to expand any Ttensor into a
Hermitian T-product tensor. For any tensor $\mathcal{C}\in\mathbb{C}^{m\times
n\times p}$, a dilation for the tensor $\mathcal{C}$, denoted as
$\mathfrak{D}(\mathcal{C})$, will be

$\displaystyle\mathfrak{D}(\mathcal{C})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\left[\begin{array}[]{cc}\mathcal{O}&\mathcal{C}\\\
\mathcal{C}^{\mathrm{H}}&\mathcal{O}\\\ \end{array}\right],$ (81)

where $\mathfrak{D}(\mathcal{C})\in\mathbb{C}^{(m+n)\times(m+n)\times p}$ and
we have
$\left(\mathfrak{D}(\mathcal{C})\right)^{\mathrm{H}}=\mathfrak{D}(\mathcal{C})$
(Hermitian T-product tensor after dilation). Also, we have

$\displaystyle\mathfrak{D}^{2}(\mathcal{C})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\left[\begin{array}[]{cc}\mathcal{C}\star\mathcal{C}^{\mathrm{H}}&\mathcal{O}\\\
\mathcal{O}&\mathcal{C}^{\mathrm{H}}\star\mathcal{C}\\\ \end{array}\right].$
(84)

Since Eq. (81) is zero trace, the largest eigenvalue of
$\mathfrak{D}(\mathcal{C})$ will be the same with the largest singular of
$\mathcal{C}$.

Since the expectation of a random T-product tensor can be considered as a
convex combination, expectation preserves the TPSD order as:

$\displaystyle\mathcal{X}\preceq\mathcal{Y}\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{almost surely}\leavevmode\nobreak\
\leavevmode\nobreak\
\Longrightarrow\mathbb{E}\mathcal{X}\preceq\mathbb{E}\mathcal{Y}.$ (85)

Also from Lemma 7, we know that the quadratic function $f(x)=x^{2}$ is
T-product tensor convex, thus, we have

$\displaystyle\left(\mathbb{E}\mathcal{C}\right)^{2}\preceq\mathbb{E}\left(\mathcal{C}^{2}\right).$
(86)

We will present one more theorem in this section about Golden-Thompson
inequality for T-product tensors. We recall 1.2

See 1.2 Proof: From T-SVD decomposition and Eq. (41), we can express the
tensor $\mathcal{C}$ as

$\displaystyle\mathcal{C}=\sum\limits_{\lambda}\mathcal{P}_{\lambda},$ (87)

where $\lambda$ are eigenvalues and $\mathcal{P}_{\lambda}$ are corresponding
projectors (T-product tensors) which are mutually orthgonal. Given
$\mathcal{X}\succeq\mathcal{O}$, we define following mapping with respect to
the tensor $\mathcal{C}$ as :

$\displaystyle\mathfrak{P}_{\mathcal{C}}(\mathcal{X}):\mathcal{X}\rightarrow\sum\limits_{\lambda}\mathcal{P}_{\lambda}\star\mathcal{X}\star\mathcal{P}_{\lambda}.$
(88)

Then, we have following properties about mapping
$\mathfrak{P}_{\mathcal{C}}(\mathcal{X})$

  1. 1.

$\mathfrak{P}_{\mathcal{C}}(\mathcal{X})$ commutes with $\mathcal{C}$;

  2. 2.

$\mathrm{Tr}\left(\mathfrak{P}_{\mathcal{C}}(\mathcal{X})\star\mathcal{C}\right)=\mathrm{Tr}\left(\mathcal{X}\star\mathcal{C}\right)$;

  3. 3.

$\mathfrak{P}_{\mathcal{C}}(\mathcal{X})\succeq\frac{\mathcal{X}}{\left|sp(\mathcal{C})\right|}$,
where
$sp(\mathcal{C})=\left\\{\lambda_{1},\lambda_{2},\cdots,\lambda_{\left|sp(\mathcal{C})\right|}\right\\}$.

The third property of $\mathfrak{P}_{\mathcal{C}}(\mathcal{X})$ is true due to
the following relation:

$\displaystyle\mathfrak{P}_{\mathcal{C}}(\mathcal{X})$ $\displaystyle=$
$\displaystyle\sum\limits_{\lambda\in
sp(\mathcal{C})}\mathcal{P}_{\lambda}\star\mathcal{X}\star\mathcal{P}_{\lambda}$
(89) $\displaystyle=$
$\displaystyle\frac{1}{\left|sp(\mathcal{C})\right|}\sum\limits_{x=1}^{\left|sp(\mathcal{C})\right|}\mathcal{U}_{x}\star\mathcal{X}\star\mathcal{U}^{\mathrm{H}}_{x}$
$\displaystyle\succeq$
$\displaystyle\frac{\mathcal{X}}{\left|sp(\mathcal{C})\right|},$

where
$\mathcal{U}_{x}=\sum\limits_{i=1}^{\left|sp(\mathcal{C})\right|}\exp\left(\frac{\sqrt{-1}2\pi
xi}{\left|sp(\mathcal{C})\right|}\right)\mathcal{P}_{\lambda_{i}}$.

Let $\mathcal{A}_{1}=\exp(\mathcal{C})$ and
$\mathcal{A}_{2}=\exp(\mathcal{D})$, we have

$\displaystyle\log\mathrm{Tr}\left(\exp\left(\log\mathcal{A}_{1}+\log\mathcal{A}_{2}\right)\right)$
$\displaystyle=_{1}$
$\displaystyle\frac{1}{n}\log\mathrm{Tr}\left(\exp\left(\log\mathcal{A}^{\otimes
n}_{1}+\log\mathcal{A}^{\otimes n}_{2}\right)\right)$ (90)
$\displaystyle\leq_{2}$
$\displaystyle\frac{1}{n}\log\mathrm{Tr}\left(\exp\left(\log\mathfrak{P}_{\mathcal{A}^{\otimes
n}_{2}}(\mathcal{A}_{1}^{\otimes n})+\log\mathcal{A}^{\otimes
n}_{2}\right)\right)$ $\displaystyle+\frac{\log\mbox{poly}(n)}{n}$
$\displaystyle=_{3}$
$\displaystyle\frac{1}{n}\log\left(\mathrm{Tr}\left(\mathfrak{P}_{\mathcal{A}^{\otimes
n}_{2}}(\mathcal{A}_{1}^{\otimes n})\star\mathcal{A}^{\otimes
n}_{2}\right)\right)+\frac{\log\mbox{poly}(n)}{n}$ $\displaystyle=_{4}$
$\displaystyle\log\mathrm{Tr}\left(\mathcal{A}_{1}\star\mathcal{A}_{2}\right)+\frac{\log\mbox{poly}(n)}{n}.$

he equality $=_{1}$ comes from the fact that the trace is multiplicative under
the Kronecker product. The inequality $\leq_{2}$ follows from inequality from
the third property of $\mathfrak{P}_{\mathcal{A}^{\otimes
m}_{2}}(\mathcal{A}_{1}^{\otimes m})$, the monotone of $\log$ and
$\mathrm{Tr}\exp\left(\leavevmode\nobreak\ \right)$ functions, and the number
of eigenvalues of $\mathcal{A}_{2}^{\otimes n}$ growing polynomially with $n$
due to the fact that the number of distinct eigenvalues of
$\mathcal{A}_{2}^{\otimes n}$ is bounded by the number of different types of
sequences of $mp$ symbols of length $n$, see Lemma II.1 in [18]. The equality
$=_{3}$ utilizes the commutativity property for tensors
$\mathfrak{P}_{\mathcal{A}^{\otimes m}_{2}}(\mathcal{A}_{1}^{\otimes m})$ and
$\mathcal{A}^{\otimes m}_{2}$ based on the first property. Finally, the
equality $=_{4}$ applies trace properties from the second property of the
mapping $\mathfrak{P}_{\mathcal{A}^{\otimes m}_{2}}(\mathcal{A}_{1}^{\otimes
m})$. If $n\rightarrow\infty$, the result of this theorem is established.
$\hfill\Box$

##  3 Lieb’s Concavity Under T-product

In this section, we will extend several trace inequalities to T-product
tensors: Jensen’s T-product tensor inequality in Section 3.1 and Klein’s
T-product tensor inequality in Section 3.2. These new T-product tensor
inequalities will play important roles in establishing a new version of Lieb’s
concavity theorem under T-product in Section 3.3.

###  3.1 Jensen’s T-product Inequality

In this subsection, we will derive Jensen’s T-product tensor inequality in
Theorem 1.3. We begin with a lemma which will be used in later proof in
Thereom 1.3.

Given two natural numbers $m,n$, we define a T-product tensor
$\theta\in\mathbb{C}^{m\times m\times p}$ as
$\exp(2\pi\sqrt{-1}/n)\times\mathcal{I}_{mmp}$. Then, we can have tensor
$\mathcal{D}\in\mathbb{C}^{mn\times mn\times p}$ obtained by

$\displaystyle\mathcal{D}=\mbox{diag}\left(\overbrace{\theta,\theta^{2},\cdots,\theta^{n-1},\mathcal{I}_{mmp}}^{\mbox{total
$n$ T-product tensors}}\right),$ (91)

where
$\mbox{diag}\left(\theta,\theta^{2},\cdots,\theta^{n-1},\mathcal{I}_{mmp}\right)$
will be a matrix with entries as T-product tensors, and the _diagonal_ part of
this matrix is compsoded by tensors
$(\theta,\theta^{2},\cdots,\theta^{n-1},\mathcal{I}_{mmp})$. Let
$\mathcal{D}\in\mathbb{C}^{mn\times mn\times p}$ be another matrix of
T-product tensor, i.e., entries $d_{i,j}$ as T-product tensors. We define a
new operation
$\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}$ between two
T-product tensors, $\mathcal{A},\mathcal{B}$ with dimensions belong to
$\mathbb{C}^{mn\times mn\times p}$ as:

$\displaystyle\left(\mathcal{A}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{B}\right)\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\sum\limits_{k=1}^{n}a_{i,k}\star b_{k,j},$ (92)

where both $a_{i,k}$ and $c_{k,j}$ are T-product tensors. Therefore, given any
tensor $\mathcal{C}\in\mathbb{C}^{mn\times mn\times p}$, the $i,j$-th entry (a
T-product tensor) of
$\mathcal{C}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{D}$
becoms $\exp(2\pi\sqrt{-1}j/n)\times c_{i,j}$, where
$c_{i,j}\in\mathbb{C}^{m\times m\times p}$ is a T-product tensor.

###### Lemma 9

Given any tensor $\mathcal{C}\in\mathbb{C}^{mn\times mn\times p}$ and the
tensor $\mathcal{D}$ defined by Eq. 91, we have

$\displaystyle\frac{1}{n}\sum\limits_{k=1}^{n}\mathcal{D}^{-k}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{C}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{D}^{k}$
$\displaystyle=$
$\displaystyle\mbox{diag}\left(c_{1,1},c_{2,2},\cdots,c_{n,n}\right)$ (93)

where $\mathcal{D}^{k}$ is the self-product of the tensor $\mathcal{D}$ by
$\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}$ operation
$k$-times.

Proof: By direct computation with
$\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}$, we have
following:

$\displaystyle\left(\frac{1}{n}\sum\limits_{k=1}^{n}\mathcal{D}^{-k}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{C}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{D}^{k}\right)_{i,j}=\frac{1}{n}\sum\limits_{k=1}^{n}\left(\exp(2\pi\sqrt{-1}(j-i)/n)\right)^{k}c_{i,j},$
(94)

where this summation is zero for $i\neq j$, otherwise, it is $c_{i,i}$.
$\hfill\Box$

We are ready to prove 1.3.

See 1.3

Proof: Let us define a unitary tensor
$\mathcal{U}=(u_{i,j})\in\mathbb{C}^{mn\times mn\times p}$ for $1\leq i,j\leq
n$ as $u_{i,j}=\mathcal{C}_{i}$,
$\mathcal{D}=\mbox{diag}\left(\theta,\theta^{2},\cdots,\theta^{n-1},\mathcal{I}_{mmp}\right)$
defined by Eq. (91), and define the tensor
$\overline{\mathcal{X}}\in\mathbb{C}^{mn\times mn\times p}$ as
$\mbox{diag}\left(\mathcal{X}_{1},\cdots,\mathcal{X}_{n}\right)$. From Lemma
9, we have

$\displaystyle
f\left(\sum\limits_{i=1}^{n}\mathcal{C}_{i}^{\mathrm{H}}\star\mathcal{X}_{i}\star\mathcal{C}_{i}\right)$
$\displaystyle=$ $\displaystyle
f\left(\left(\mathcal{U}^{\mathrm{H}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\overline{\mathcal{X}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}\right)_{n,n}\right)$
(95) $\displaystyle=$ $\displaystyle
f\left(\left(\sum\limits_{i=1}^{n}\frac{1}{n}\mathcal{D}^{-i}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}^{\mathrm{H}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\overline{\mathcal{X}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{D}^{i}\right)_{n,n}\right)$
$\displaystyle=$ $\displaystyle
f\left(\left(\sum\limits_{i=1}^{n}\frac{1}{n}\mathcal{D}^{-i}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}^{\mathrm{H}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\overline{\mathcal{X}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{D}^{i}\right)\right)_{n,n}$
$\displaystyle\leq$
$\displaystyle\left(\frac{1}{n}\sum\limits_{i=1}^{n}f\left(\mathcal{D}^{-i}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}^{\mathrm{H}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\overline{\mathcal{X}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{D}^{i}\right)\right)_{n,n}$
$\displaystyle=$
$\displaystyle\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathcal{D}^{-i}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}^{\mathrm{H}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}f\left(\overline{\mathcal{X}}\right)\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{D}^{i}\right)_{n,n}$
$\displaystyle=$
$\displaystyle\left(\mathcal{U}^{\mathrm{H}}\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}f\left(\overline{\mathcal{X}}\right)\stackMath\mathbin{\stackinset{c}{0ex}{c}{0ex}{\ast}{\bigcirc}}\mathcal{U}\right)_{n,n}$
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{n}\mathcal{C}_{i}^{\mathrm{H}}\star
f(\mathcal{X}_{i})\star\mathcal{C}_{i},$

where the inequality comes from that the function $f$ is a _T-product tensor
convex_ function. $\hfill\Box$

###  3.2 Klein’s T-product Inequality

The immediate application of Theorem 1.1 is to prove Klein’s inequality for
T-product tensor. We recall 1.4.

See 1.4

Proof: We define function $F(t)$ as

$\displaystyle
F(t)=\mathrm{Tr}\left(f\left(\mathcal{D}+t\left(\mathcal{C}-\mathcal{D}\right)\right)\right),$
(96)

where $t\in(0,1)$. From Theorem 1.1, $F(t)$ is a convex function. Then, we
have

$\displaystyle F(0)+t(F(1)-F(0))\geq F(t)\Longleftrightarrow
F(1)-F(0)\geq\frac{F(t)-F(0)}{t}$ (97)

By taking limit $t\rightarrow 0$ at $F(1)-F(0)\geq\frac{F(t)-F(0)}{t}$, we
have

$\displaystyle F(1)-F(0)\geq F^{\prime}(0),$ (98)

then we obatin Klein’s ineqaulity under T-product by rearrangement and
substitution with Eq. (96). $\hfill\Box$

###  3.3 Lieb’s Concavity Theorem Under T-product

In this section, we will extend Lieb’s concavity theorem to T-product tensors
and we begin with the definition about the relative entropy between two
T-product tensors.

###### Definition 3

Given two TPD tensors $\mathcal{A}\in\mathbb{C}^{m\times m\times p}$ and
tensor $\mathcal{B}\in\mathbb{C}^{m\times m\times p}$. The relative entropy
between two T-product tensors $\mathcal{A}$ and $\mathcal{B}$ is defined as

$\displaystyle
D(\mathcal{A}\parallel\mathcal{B})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathrm{Tr}\mathcal{A}\star(\log\mathcal{A}-\log\mathcal{B}).$ (99)

We apply _perspective function_ concept for T-product tensor convex and
introduce the following lemma about the convexity of a T-product tensor convex
function [19].

###### Lemma 10

Given $f$ as a convex function, two commuting tensors
$\mathcal{X},\mathcal{Y}\in\mathbb{C}^{m\times m\times p}$, i.e.,
$\mathcal{X}\star\mathcal{Y}=\mathcal{Y}\star\mathcal{X}$, and the existence
of the $\mathcal{Y}^{-1}$, then the following map $h$

$\displaystyle
h(\mathcal{X},\mathcal{Y})=f(\mathcal{X}\star\mathcal{Y}^{-1})\star\mathcal{Y}$
(100)

is jointly convex in the sense that, given $t\in[0,1]$, if
$\mathcal{X}=t\mathcal{X}_{1}+(1-t)\mathcal{X}_{2}$ and
$\mathcal{Y}=t\mathcal{Y}_{1}+(1-t)\mathcal{Y}_{2}$ with
$\mathcal{X}_{1}\star\mathcal{Y}_{1}=\mathcal{Y}_{1}\star\mathcal{X}_{1}$ and
$\mathcal{X}_{2}\star\mathcal{Y}_{2}=\mathcal{Y}_{2}\star\mathcal{X}_{2}$, we
should have

$\displaystyle h(\mathcal{X},\mathcal{Y})\leq
th(\mathcal{X}_{1},\mathcal{Y}_{1})+(1-t)h(\mathcal{X}_{2},\mathcal{Y}_{2}).$
(101)

###### Proof 1

Constructing tensors
$\mathcal{A}=(t\mathcal{Y}_{1})^{1/2}\star\mathcal{Y}^{-1/2}$ and
$\mathcal{B}=((1-t)\mathcal{Y}_{2})^{1/2}\star\mathcal{Y}^{-1/2}$, then we
have

$\displaystyle\mathcal{A}^{\mathrm{H}}\star\mathcal{A}+\mathcal{B}^{\mathrm{H}}\star\mathcal{B}=\mathcal{I}_{mmp}$
(102)

Since we have

$\displaystyle h(\mathcal{X},\mathcal{Y})$ $\displaystyle=$ $\displaystyle
f(\mathcal{X}\star\mathcal{Y}^{-1})\star\mathcal{Y}$ (103) $\displaystyle=$
$\displaystyle\mathcal{Y}^{1/2}\star
f(\mathcal{Y}^{-1/2}\star\mathcal{X}\star\mathcal{Y}^{-1/2})\star\mathcal{Y}^{1/2}$
$\displaystyle=$ $\displaystyle\mathcal{Y}^{1/2}\star
f(\mathcal{A}^{\mathrm{H}}\star\mathcal{X}_{1}\star\mathcal{Y}^{-1}_{1}\star\mathcal{A}+\mathcal{B}^{\mathrm{H}}\star\mathcal{X}_{2}\star\mathcal{Y}^{-1}_{2}\star\mathcal{B})\star\mathcal{Y}^{1/2}$
$\displaystyle\leq_{1}$
$\displaystyle\mathcal{Y}^{1/2}\star\left(\mathcal{A}^{\mathrm{H}}\star
f(\mathcal{X}_{1}\star\mathcal{Y}^{-1}_{1})\star\mathcal{A}\right.$
$\displaystyle\left.+\mathcal{B}^{\mathrm{H}}\star
f(\mathcal{X}_{2}\star\mathcal{Y}^{-1}_{2})\star\mathcal{B}\right)\star\mathcal{Y}^{1/2}$
$\displaystyle=$
$\displaystyle(t\mathcal{Y}_{1})^{1/2}f(\mathcal{X}_{1}\star\mathcal{Y}^{-1}_{1})(t\mathcal{Y}_{1})^{1/2}+((1-t)\mathcal{Y}_{2})^{1/2}f(\mathcal{X}_{2}\star\mathcal{Y}^{-1}_{2})((1-t)\mathcal{Y}_{2})^{1/2}$
$\displaystyle=$ $\displaystyle
th(\mathcal{X}_{1},\mathcal{Y}_{1})+(1-t)h(\mathcal{X}_{2},\mathcal{Y}_{2})$

where $\leq_{1}$ is based on the condition provided by Eq. (102) and Theorem
1.3.

Following lemma is given to establish the joint convexity property of relative
entropy for T-product tensors.

######  Lemma 11 (Joint Convexity of Relative Entropy for T-product Tensors)

The relative entropy function of two TPD tensors is a jointly convex function.
That is

$\displaystyle\mathbb{D}(t\mathcal{A}_{1}+(1-t)\mathcal{A}_{2}\parallel
t\mathcal{B}_{1}+(1-t)\mathcal{B}_{2})\leq
t\mathbb{D}(\mathcal{A}_{1}\parallel\mathcal{B}_{1})+(1-t)\mathbb{D}(\mathcal{A}_{2}\parallel\mathcal{B}_{2}),$
(104)

where $t\in[0,1]$ and all the following four tensors $\mathcal{A}_{1}$,
$\mathcal{B}_{1}$, $\mathcal{A}_{2}$ and $\mathcal{B}_{2}$, are TPD tensors.

Proof: From the definition 3, we wish to show the joint convexity of the
function $\mathbb{D}(\mathcal{A}\parallel\mathcal{B})$ with respect to the
tensors $\mathcal{A},\mathcal{B}\in\mathbb{C}^{m\times m\times p}$. Let us
define tensor operators
$\mathcal{F}(\mathcal{X})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathcal{A}\star\mathcal{X}$ and
$\mathcal{G}(\mathcal{X})\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathcal{X}\star\mathcal{B}$ for the variable tensor
$\mathcal{X}\in\mathbb{C}^{m\times m\times p}$. Then, we have
$\mathcal{F}(\mathcal{X})$ and $\mathcal{G}(\mathcal{X})$ commuting on the
inner product operation
$\langle\mathcal{F}(\mathcal{X}),\mathcal{G}(\mathcal{X})\rangle$ defined as:

$\displaystyle\langle\mathcal{F}(\mathcal{X}),\mathcal{G}(\mathcal{X})\rangle$
$\displaystyle=$
$\displaystyle\mathrm{Tr}(\mathcal{F}^{\mathrm{H}}(\mathcal{X})\star\mathcal{G}(\mathcal{X}))$
(105)

Then, we have
$\mathrm{Tr}(\mathcal{F}^{\mathrm{H}}(\mathcal{X})\star\mathcal{G}(\mathcal{X}))=\mathrm{Tr}(\mathcal{G}^{\mathrm{H}}(\mathcal{X})\star\mathcal{F}(\mathcal{X}))$.
Since the function $f(x)=x\log x$ is tensor convex, we apply Lemma 10 to
operators $\mathcal{F}(\leavevmode\nobreak\ \leavevmode\nobreak\
),\mathcal{G}(\leavevmode\nobreak\ \leavevmode\nobreak\ )$ and the function
$h$ definition provided by Eq. (100) to obtain the following relation
($\mathcal{I}=\mathcal{I}_{mmp}$ in this proof):

$\displaystyle\langle\mathcal{I},h(\mathcal{F}(\mathcal{I}),\mathcal{G}(\mathcal{I}))\rangle$
$\displaystyle=$ $\displaystyle\langle\mathcal{I},\leavevmode\nobreak\
\leavevmode\nobreak\
\mathcal{G}(\mathcal{I})\star(\mathcal{F}(\mathcal{I})\star\mathcal{G}^{-1}(\mathcal{I}))\log(\mathcal{F}(\mathcal{I})\star\mathcal{G}^{-1}(\mathcal{I}))\rangle$
(106) $\displaystyle=$
$\displaystyle\langle\mathcal{I},\mathcal{F}(\mathcal{I})(\log\mathcal{F}(\mathcal{I})-\log\mathcal{G}(\mathcal{I}))\rangle$
$\displaystyle=$
$\displaystyle\mathrm{Tr}(\mathcal{A}\log\mathcal{A}-\mathcal{A}\log\mathcal{B})=\mathbb{D}(\mathcal{A}\parallel\mathcal{B}),$

is jointly convex with respect to tensors $\mathcal{A}$ and $\mathcal{B}$.
$\hfill\Box$

Lieb’s concavity theorem is recalled below by 1.5.

See 1.5

Proof: From Klein’s inequality obtain from Theorem 1.4, the convexity of map
$t\rightarrow t\log t$ (which is strictly concave for $t>0$) and Hermitian T-
tensors $\mathcal{X},\mathcal{Y}$, we have

$\displaystyle\mathrm{Tr}\mathcal{Y}\geq\mathrm{Tr}\mathcal{X}-\mathrm{Tr}\mathcal{X}\log\mathcal{X}+\mathrm{Tr}\mathcal{X}\log\mathcal{Y}.$
(107)

If we replace $\mathcal{Y}$ by $e^{\mathcal{H}+\log\mathcal{A}}$, we then have

$\displaystyle\mathrm{Tr}e^{\mathcal{H}+\log\mathcal{A}}=\max\limits_{\mathcal{X}\succ\mathcal{O}}\Big{\\{}\mathrm{Tr}\mathcal{X}\star\mathcal{H}-\mathbb{D}(\mathcal{X}\parallel\mathcal{A})+\mathrm{Tr}\mathcal{X}\Big{\\}}$
(108)

where $\mathbb{D}(\mathcal{X}\parallel\mathcal{A})$ is the quantum relative
entropy between two tensor operators. For real number $t\in[0,1]$ and two
positive-definite tensors $\mathcal{A}_{1},\mathcal{A}_{2}$, we have

$\displaystyle\mathrm{Tr}e^{\mathcal{H}+\log(t\mathcal{A}_{1}+(1-t)\mathcal{A}_{2})}$
$\displaystyle=$
$\displaystyle\max_{\mathcal{X}\succ\mathcal{O}}\Big{\\{}\mathrm{Tr}\mathcal{X}\mathcal{H}-\mathbb{D}(\mathcal{X}\parallel
t\mathcal{A}_{1}+(1-t)\mathcal{A}_{2})+\mathrm{Tr}\mathcal{X}\Big{\\}}$ (109)
$\displaystyle\geq$ $\displaystyle
t\max_{\mathcal{X}\succ\mathcal{O}}\Big{\\{}\mathrm{Tr}\mathcal{X}\mathcal{H}-\mathbb{D}(\mathcal{X}\parallel
t\mathcal{A}_{1})+\mathrm{Tr}\mathcal{X}\Big{\\}}$
$\displaystyle+(1-t)\max_{\mathcal{X}\succ\mathcal{O}}\Big{\\{}\mathrm{Tr}\mathcal{X}\mathcal{H}-\mathbb{D}(\mathcal{X}\parallel(1-t)\mathcal{A}_{2})+\mathrm{Tr}\mathcal{X}\Big{\\}}$
$\displaystyle=$ $\displaystyle
t\mathrm{Tr}e^{\mathcal{H}+\log\mathcal{A}_{1}}+(1-t)\mathrm{Tr}e^{\mathcal{H}+\log\mathcal{A}_{2}},$

where the first and last equalities are obtained based on the variational
formula provided by Eq. (108), and the inequality is due to the joint
convexity property of the relative entropy from Leamm 11. $\hfill\Box$

Based on Lieb’s concavity theorem for T-product tensors, we have the following
corollary.

###### Corollary 2

Let $\mathcal{A}$ be a fixed Hermitian T-product tensor, and let $\mathcal{X}$
be a random Hermitian T-product tensor, then we have

$\displaystyle\mathbb{E}\mathrm{Tr}e^{\mathcal{A}+\mathcal{X}}\leq\mathrm{Tr}e^{\mathcal{A}+\log\left(\mathbb{E}e^{\mathcal{X}}\right)}.$
(110)

###### Proof 2

Define the random tensor $\mathcal{Y}=e^{\mathcal{X}}$, we have

$\displaystyle\mathbb{E}\mathrm{Tr}e^{\mathcal{A}+\mathcal{X}}=\mathbb{E}\mathrm{Tr}e^{\mathcal{A}+\log\mathcal{Y}}\leq\mathrm{Tr}e^{\mathcal{A}+\log\left(\mathbb{E}\mathcal{Y}\right)}=\mathrm{Tr}e^{\mathcal{A}+\log\left(\mathbb{E}e^{\mathcal{X}}\right)},$
(111)

where the inequality is based on Lieb’s concavity theorem for T-product
tensors obtained by Theorem 1.5 and Jensen’s T-product tensor inequality by
Theorem 1.3.

###  3.4 T-product Tensor Moments and Cumulants

Since the expectation of a random T-product tensor can be treated as convex
combination, expectation will preserve the semidefinite order as

$\displaystyle\mathcal{X}\succ\mathcal{Y}\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ almost surely}\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \Rightarrow\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \mathbb{E}(\mathcal{X})\succ\mathbb{E}(\mathcal{Y}).$
(112)

From Jensen’s T-product tensor inequality by Theorem 1.3, we also have

$\displaystyle\mathbb{E}(\mathcal{X}^{2})\succeq\left(\mathbb{E}(\mathcal{X})\right)^{2}.$
(113)

Suppose a random Hermitian T-product tensor $\mathcal{X}$ having tensor
moments of all orders, i.e., $\mathbb{E}(\mathcal{X}^{n})$ existing for all
$n$, we can define the tensor moment-generating function, denoted as
$\mathbb{M}_{\mathcal{X}}(t)$, and the tensor cumulant-generating function,
denoted as $\mathbb{K}_{\mathcal{X}}(t)$, for the tensor $\mathcal{X}$ as

$\displaystyle\mathbb{M}_{\mathcal{X}}(t)\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\mathbb{E}e^{t\mathcal{X}},\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ and\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\
}\mathbb{K}_{\mathcal{X}}(t)\stackrel{{\scriptstyle\mbox{\tiny
def}}}{{=}}\log\mathbb{E}e^{t\mathcal{X}},$ (114)

where $t\in\mathbb{R}$. Both the tensor moment-generating function and the
tensor cumulant-generating function can be expressed as power series
expansions:

$\displaystyle\mathbb{M}_{\mathcal{X}}(t)=\mathcal{I}+\sum\limits_{n=1}^{\infty}\frac{t^{n}}{n!}\mathbb{E}(\mathcal{X}^{n}),\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ and\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\
}\mathbb{K}_{\mathcal{X}}(t)=\sum\limits_{n=1}^{\infty}\frac{t^{n}}{n!}\psi_{n},$
(115)

where $\psi_{n}$ is called _tensor cumulant_. The tensor cumulant $\psi_{n}$
can be expressed as a polynomial in terms of tensor moments up to the order
$n$, for example, the first cumulant is the mean and the second cumulant is
the variance:

$\displaystyle\psi_{1}=\mathbb{E}(\mathcal{X}),\mbox{\leavevmode\nobreak\
\leavevmode\nobreak\ and\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\
}\psi_{2}=\mathbb{E}(\mathcal{X}^{2})-(\mathbb{E}(\mathcal{X}))^{2}.$ (116)

Finally, in this work, we also assume that all random variables are
sufficiently regular for us to compute their expectations, interchange limits,
etc.

##  4 Tail Bounds By Concatenation of Lieb’s Concavity

The goal of this section is to develop several important tools which will be
applied intensively in the proof of probability inequalities for the extreme
eigentule (or eigenvalue) of a sum of independent random T-product tensors.
The first tool is the Laplace transform method for T-product tensors discussed
in Section 4.1, and the second tool is the tail bound for independent sums of
random Hermitian T-product tensors presented by Section 4.2.

###  4.1 Laplace Transform Method for T-product Tensors

We extend the Laplace transform bound from matrices to T-product tensors based
on [20]. Following lemma is given to establish the Laplace transform bound for
the maximum eigenvalue of a T-product tensor.

######  Lemma 12 (Laplace Transform Method for T-product Tensors: Eigenvalue
Version)

Let $\mathcal{X}$ be a random Hermitian T-product tensor. For
$\theta\in\mathbb{R}$, we have

$\displaystyle\mathbb{P}(\lambda_{\max}(\mathcal{X})\geq\theta)\leq\inf_{t>0}\Big{\\{}e^{-\theta
t}\mathbb{E}\mathrm{Tr}e^{t\mathcal{X}}\Big{\\}}$ (117)

###### Proof 3

Given a fix value $t$, we have

$\displaystyle\mathbb{P}(\lambda_{\max}(\mathcal{X})\geq\theta)=\mathbb{P}(\lambda_{\max}(t\mathcal{X})\geq
t\theta)=\mathbb{P}(e^{\lambda_{\max}(t\mathcal{X})}\geq e^{t\theta})\leq
e^{-t\theta}\mathbb{E}e^{\lambda_{\max}(t\mathcal{X})}.$ (118)

The first equality uses the homogeneity of the maximum eigenvalue map, the
second equality comes from the monotonicity of the scalar exponential
function, and the last relation is Markov’s inequality. Because we have

$\displaystyle
e^{\lambda_{\max}(t\mathcal{Y})}=\lambda_{\max}(e^{t\mathcal{Y}})\leq\mathrm{Tr}e^{t\mathcal{Y}},$
(119)

where the first equality used the spectral mapping theorem from Lemma 2, and
the inequality holds because the exponential of an Hermitian T-product tensor
is TPD and the maximum eigenvalue of a TPD tensor is dominated by the trace
from Eq. (28). From Eqs (118) and (119), this lemma is established.

The Lemma 12 helps us to control the tail probabilities for the maximum
eigenvalue of a random Hermitian T-product tensor by utilizing a bound for the
trace of the tensor moment-generating function introduced in Section 3.4.

Since T-product tensors also have notions about eigentuples, we then extend
Lemma 12 from eigenvalues version to eigentuples version. We begin with the
derivation of Markov’s inequality for random vectors.

######  Lemma 13 (Markov’s inequality for Random Vector)

If $\mathbf{X}\in\mathbb{R}^{p}$ is a nonnegative random vector and
$\mathbf{a}>\mathbf{0}$, then the probability that $\mathbf{X}$ is at least
$\mathbf{a}=[a_{i}]$ can be bounded as:

$\displaystyle\mathrm{Pr}\left(\mathbf{X}\geq\mathbf{a}\right)\leq\min\limits_{i}\left\\{\frac{\left(\mathbb{E}\left(\mathbf{X}\right)\right)_{i}}{a_{i}}\right\\}$
(120)

where $1\leq i\leq p$.

Proof: Because we have

$\displaystyle\mathbb{E}\left(\mathbf{X}\right)$ $\displaystyle=$
$\displaystyle\int\limits_{\mathbf{0}}^{\infty^{p}}\mathbf{x}f(\mathbf{x})d\mathbf{x}=\int\limits_{\mathbf{0}}^{\mathbf{a}}\mathbf{x}f(\mathbf{x})d\mathbf{x}+\int\limits_{\mathbf{a}}^{\infty^{p}}\mathbf{x}f(\mathbf{x})d\mathbf{x}$
(121) $\displaystyle\geq$
$\displaystyle\int\limits_{\mathbf{a}}^{\infty^{p}}\mathbf{x}f(\mathbf{x})d\mathbf{x}\geq\int\limits_{\mathbf{a}}^{\infty^{p}}\mathbf{a}f(\mathbf{x})d\mathbf{x}=\mathbf{a}\int\limits_{\mathbf{a}}^{\infty^{p}}f(\mathbf{x})d\mathbf{x}$
$\displaystyle=$
$\displaystyle\mathbf{a}\mathrm{Pr}\left(\mathbf{X}\geq\mathbf{a}\right),$

therefore, we have the desired inequality shown by Eq. (120). $\hfill\Box$

We are ready to present following lemma about Laplace transform method for
T-product tensors with eigentuples.

######  Lemma 14 (Laplace Transform Method for T-product Tensors: Eigentuple
Version)

Let $\mathcal{X}\in\mathbb{C}^{m\times m\times p}$ be a random T-positive
definite (TPD) tensor and an all one vector
$\mathbf{1}_{p}=[1,1,\cdots,1]^{\mathrm{T}}\in\mathbb{C}^{p}$. Suppose we have

$\displaystyle\frac{1}{p}\lambda_{\max}^{p}(e^{t\mathcal{X}})+1-\frac{1}{p}\leq\mathrm{Tr}(e^{t\mathcal{X}}),$
(122)

where $t>0$ 111If we scale the random TPD tensor $\mathcal{X}$ as the
$\lambda_{\max}(e^{t\mathcal{X}})=1$, then Eq. (122) is always valid.. Then,
for $\mathbf{b}\in\mathbb{R}^{p}$, we obtain

$\displaystyle\mathbb{P}(\mathbf{d}_{\max}(\mathcal{X})\geq\mathbf{b})\leq\inf_{t>0}\min\limits_{i}\left\\{\frac{\mathbb{E}\left(\mathrm{Tr}\left(e^{t\mathcal{X}}\right)\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{i}}\right\\},$
(123)

where $\mathbf{d}_{\max}$ is the maximum eigentuple of the TPD tensor
$\mathcal{X}$.

Proof: Given a fix value $t$, we have

$\displaystyle\mathbb{P}(\mathbf{d}_{\max}(\mathcal{X})\geq\mathbf{b})=\mathbb{P}(\mathbf{d}_{\max}(t\mathcal{X})\geq
t\mathbf{b})=\mathbb{P}(e_{\bigodot}^{\mathbf{d}_{\max}(t\mathcal{X})}\geq
e_{\bigodot}^{t\mathbf{b}})\leq\min\limits_{i}\left\\{\frac{\left(\mathbb{E}\left(e_{\bigodot}^{\mathbf{d}_{\max}(t\mathcal{X})}\right)\right)_{i}}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{i}}\right\\}.$
(124)

The first equality uses the homogeneity of the maximum eigenvalue map, the
second equality comes from the monotonicity of the exponential function with
operation $\bigodot$ defined in Proposition 2.1 from work [1], and the last
relation is Markov’s inequality for random vector obtained from Lemma 13 since
both $\mathbb{E}\left(e_{\bigodot}^{\mathbf{d}_{\max}(t\mathcal{X})}\right)$
and $e_{\bigodot}^{t\mathbf{b}}$ are vectors with $p$ entries. Then, we have

$\displaystyle e_{\bigodot}^{\mathbf{d}_{\max}(t\mathcal{X})}\leq
e_{\bigodot}^{\lambda_{\max}(t\mathcal{X})\mathbf{1}_{p}}\leq\mathrm{Tr}\left(e^{t\mathcal{X}}\right)\mathbf{1}_{p},$
(125)

where the first inequality comes from the relation that
$\mathbf{d}_{\max}(t\mathcal{X})\leq\lambda_{\max}(t\mathcal{X})\mathbf{1}_{p}$,
and the second inequality holds because
$e^{\lambda_{\max}(t\mathcal{X})}=\lambda_{\max}(e^{t\mathcal{X}})$ and the
relation
$\frac{1}{p}\lambda_{\max}^{p}(e^{t\mathcal{X}})+1-\frac{1}{p}\leq\mathrm{Tr}(e^{t\mathcal{X}})$.
From Eqs (124) and (125), this lemma is established. $\hfill\Box$

###  4.2 Tail Bounds for Independent Sums of Random T-product Tensors

This section will present the tail bound for the sum of independent random
T-product tensors and several corollaries according to this tail bound for
independent sums. We begin with the subadditivity lemma of tensor cumulant-
generating functions.

###### Lemma 15

Given a finite sequence of independent random Hermitian T-product tensors
$\\{\mathcal{X}_{i}\\}$, where $\mathcal{X}_{i}\in\mathbb{C}^{m\times m\times
p}$, we have

$\displaystyle\mathbb{E}\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n}t\mathcal{X}_{i}\right)\leq\mathrm{Tr}\exp\left(\sum\limits_{i}^{n}\log\mathbb{E}e^{t\mathcal{X}_{i}}\right),\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for $t\in\mathbb{R}$.}$ (126)

Proof: We begin with the following definition for the tensor cumulant-
generating function for $\mathcal{X}_{i}$ as:

$\displaystyle\mathbb{K}_{i}(t)$
$\displaystyle\stackrel{{\scriptstyle\mbox{\tiny def}}}{{=}}$
$\displaystyle\log(\mathbb{E}e^{t\mathcal{X}_{i}}).$ (127)

Then, we define the Hermitian T-product tensor $\mathcal{H}_{k}$ as

$\displaystyle\mathcal{H}_{k}(t)=\sum\limits_{i=1}^{k-1}t\mathcal{X}_{i}+\sum\limits_{i=k+1}^{n}\mathbb{K}_{i}(t).$
(128)

By applying Eq. (128) to Theorem 1.5 repeatedly for $k=1,2,\cdots,n$, we have

$\displaystyle\mathbb{E}\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n}t\mathcal{X}_{i}\right)$
$\displaystyle=_{1}$
$\displaystyle\mathbb{E}_{0}\cdots\mathbb{E}_{n-1}\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n-1}t\mathcal{X}_{i}+t\mathcal{X}_{n}\right)$
$\displaystyle\leq$
$\displaystyle\mathbb{E}_{0}\cdots\mathbb{E}_{n-2}\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n-1}t\mathcal{X}_{i}+\log\left(\mathbb{E}_{n-1}e^{t\mathcal{X}_{n}}\right)\right)$
$\displaystyle=$
$\displaystyle\mathbb{E}_{0}\cdots\mathbb{E}_{n-2}\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n-2}t\mathcal{X}_{i}+t\mathcal{X}_{n-1}+\mathbb{K}_{n}(t).\right)$
$\displaystyle\leq$
$\displaystyle\mathbb{E}_{0}\cdots\mathbb{E}_{n-3}\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n-2}t\mathcal{X}_{i}+\mathbb{K}_{n-1}(t)+\mathbb{K}_{n}(t)\right)$
$\displaystyle\cdots$ $\displaystyle\leq$
$\displaystyle\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n}\mathbb{K}_{i}(t)\right)$
(129)

where the equality $=_{1}$ is based on the law of total expectation by
defining $\mathbb{E}_{i}$ as the conditional expectation given
$\mathcal{X}_{1},\cdots,\mathcal{X}_{i}$. $\hfill\Box$

We are ready to present the theorem for the tail bound of independent sums of
random Hermitian T-product tensors with respect to the maximum eigenvalue. We
recell 1.6

See 1.6

Proof: By substituting the Lemma 15 into the Laplace transform bound provided
by the Lemma 12, this theorem is established. $\hfill\Box$

Several useful corollaries will be provided based on Theorem 1.6.

###### Corollary 3

Given a finite sequence of independent Hermitian random tensors
$\\{\mathcal{X}_{i}\\}\in\mathbb{C}^{m\times m\times p}$. If there is a
function $f:(0,\infty)\rightarrow[0,\infty]$ and a sequence of non-random
Hermitian T-product tensors $\\{\mathcal{A}_{i}\\}$ with following condition:

$\displaystyle
f(t)\mathcal{A}_{i}\succeq\log\mathbb{E}e^{t\mathcal{X}_{i}},\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for $t>0$.}$ (130)

Then, for all $\theta\in\mathbb{R}$, we have

$\displaystyle\mathrm{Pr}\left(\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\theta\right)$
$\displaystyle\leq$ $\displaystyle
mp\inf\limits_{t>0}\Big{\\{}\exp\left[-t\theta+f(t)\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{A}_{i}\right)\right]\Big{\\}}$
(131)

Proof: From the condition provided by Eq. (130) and Theorem 1.6, we have

$\displaystyle\mathrm{Pr}\left(\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\theta\right)$
$\displaystyle\leq$ $\displaystyle
e^{-t\theta}\mathrm{Tr}\exp(f(t)\sum\limits_{i=1}^{n}\mathcal{A}_{i})$ (132)
$\displaystyle\leq$ $\displaystyle
mpe^{-t\theta}\lambda_{\max}\left(\exp(f(t)\sum\limits_{i=1}^{n}\mathcal{A}_{i})\right)$
$\displaystyle=$ $\displaystyle
mpe^{-t\theta}\exp\left(f(t)\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{A}_{i}\right)\right),$

where the second inequality holds since we bound the trace of a TPD T-product
tensor by the dimension size $m\times p$ multiplied by the maximum eigenvalue;
the last equality is based on the spectral mapping theorem since the function
$f$ is nonnegative. $\hfill\Box$

###### Corollary 4

Given a finite sequence of independent Hermitian random tensors
$\\{\mathcal{X}_{i}\\}\in\mathbb{C}^{m\times m\times p}$. For all
$\theta\in\mathbb{R}$, we have

$\displaystyle\mathrm{Pr}\left(\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\theta\right)$
$\displaystyle\leq$ $\displaystyle
mp\inf\limits_{t>0}\Big{\\{}\exp\left[-t\theta+n\log\lambda_{\max}\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathbb{E}e^{t\mathcal{X}_{i}}\right)\right]\Big{\\}}$

Proof: From T-tensor logarithm concavity property provided by Lemma 8, we have

$\displaystyle\sum\limits_{i=1}^{n}\log\mathbb{E}e^{t\mathcal{X}_{i}}=n\cdot\frac{1}{n}\sum\limits_{i=1}^{n}\log\mathbb{E}e^{t\mathcal{X}_{i}}\preceq
n\log\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathbb{E}e^{t\mathcal{X}_{i}}\right),$
(134)

and from the trace exponential monotone property provided by Lemma 3, we have

$\displaystyle\mathrm{Pr}\left(\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\theta\right)\leq
e^{-t\theta}\mathrm{Tr}\exp\left(n\log\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathbb{E}e^{t\mathcal{X}_{i}}\right)\right)\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ $ $\displaystyle\leq
mp\inf\limits_{t>0}\Big{\\{}\exp\left[-t\theta+n\log\lambda_{\max}\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathbb{E}e^{t\mathcal{X}_{i}}\right)\right]\Big{\\}},$
(135)

where the last inequality holds since we bound the trace of a positive-
definite tensor by the dimension size $m\times p$ multiplied by the maximum
eigenvalue and apply spectral mapping theorem twice. $\hfill\Box$

Similarly, we can generalize master tail bound for independent sum of random
Hermitian T-product tensors for eigenvalue version from Theorem 1.6 to master
tail bound for independent sum of random Hermitian T-product tensors for
eigentuple version by the following Theorem 1.7.

See 1.7

Proof: By substituting the Lemma 15 into the Laplace transform bound provided
by the Lemma 14, this theorem is established. $\hfill\Box$

Some useful corollaries will be provided based on Theorem 1.7.

###### Corollary 5

Given a finite sequence of independent random Hermitian T-product tensors
$\\{\mathcal{X}_{i}\\}$ with dimensions in $\mathbb{C}^{m\times m\times p}$.
If there is a function $f:(0,\infty)\rightarrow[0,\infty]$ and a sequence of
non-random Hermitian T-product tensors $\\{\mathcal{A}_{i}\\}$ with following
condition:

$\displaystyle
f(t)\mathcal{A}_{i}\succeq\log\mathbb{E}e^{t\mathcal{X}_{i}},\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{for $t>0$.}$ (136)

Then, for all $\mathbf{b}\in\mathbb{R}^{p}$ and
$\sum\limits_{i=1}^{n}t\mathcal{X}_{i}$ satisfing Eq. (122), we have

$\displaystyle\mathrm{Pr}\left(\mathbf{d}_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\mathbf{b}\right)$
$\displaystyle\leq$ $\displaystyle mp\inf\limits_{t>0}\min\limits_{1\leq j\leq
p}\left\\{\frac{\exp\left(f(t)\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{A}_{i}\right)\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\}.$
(137)

Proof: From the condition provided by Eq. (136) and Theorem 1.7, we have

$\displaystyle\mathrm{Pr}\left(\mathbf{d}_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\mathbf{b}\right)$
$\displaystyle\leq$ $\displaystyle\inf\limits_{t>0}\min\limits_{1\leq j\leq
p}\left\\{\frac{\mathrm{Tr}\exp\left(f(t)\sum\limits_{i=1}^{n}\mathcal{A}_{i}\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\}$
(138) $\displaystyle\leq$ $\displaystyle mp\inf\limits_{t>0}\min\limits_{1\leq
j\leq
p}\left\\{\frac{\lambda_{\max}\left(\exp\left(f(t)\sum\limits_{i=1}^{n}\mathcal{A}_{i}\right)\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\}$
$\displaystyle=$ $\displaystyle mp\inf\limits_{t>0}\min\limits_{1\leq j\leq
p}\left\\{\frac{\exp\left(f(t)\lambda_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{A}_{i}\right)\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\}$

where the second inequality holds since we bound the trace of a TPD T-tensor
by the eigenvalue size with $m\times p$ multiplied by the maximum eigenvalue;
the last equality is based on the spectral mapping theorem since the function
$f$ is nonnegative. This corollary is proved. $\hfill\Box$

###### Corollary 6

Given a finite sequence of independent random Hermitian T-product tensors
$\\{\mathcal{X}_{i}\\}$ with dimensions in $\mathbb{C}^{m\times m\times p}$, a
real vector $\mathbf{b}\in\mathbb{R}^{p}$ and
$\sum\limits_{i=1}^{n}t\mathcal{X}_{i}$ satisfing Eq. (122), we have

$\displaystyle\mathrm{Pr}\left(\mathbf{d}_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\mathbf{b}\right)$
$\displaystyle\leq$ $\displaystyle mp\inf\limits_{t>0}\min\limits_{1\leq j\leq
p}\left\\{\frac{\exp\left(n\log\lambda_{\max}\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathbb{E}e^{t\mathcal{X}_{i}}\right)\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\}$
(139)

Proof: From T-tensor logarithm concavity property provided by Lemma 8, we have

$\displaystyle\sum\limits_{i=1}^{n}\log\mathbb{E}e^{t\mathcal{X}_{i}}=n\cdot\frac{1}{n}\sum\limits_{i=1}^{n}\log\mathbb{E}e^{t\mathcal{X}_{i}}\preceq
n\log\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathbb{E}e^{t\mathcal{X}_{i}}\right),$
(140)

and from the trace exponential monotone property provided by Lemma 3, we also
have

$\displaystyle\mathrm{Pr}\left(\mathbf{d}_{\max}\left(\sum\limits_{i=1}^{n}\mathcal{X}_{i}\right)\geq\mathbf{b}\right)$
$\displaystyle\leq$ $\displaystyle\inf\limits_{t>0}\min\limits_{1\leq j\leq
p}\left\\{\frac{\mathrm{Tr}\exp\left(\sum\limits_{i=1}^{n}\log\mathbb{E}e^{t\mathcal{X}_{i}}\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\}$
(141) $\displaystyle\leq$ $\displaystyle\inf\limits_{t>0}\min\limits_{1\leq
j\leq
p}\left\\{\frac{\mathrm{Tr}\exp\left(n\log\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathbb{E}e^{t\mathcal{X}_{i}}\right)\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\}$
$\displaystyle\leq$ $\displaystyle mp\inf\limits_{t>0}\min\limits_{1\leq j\leq
p}\left\\{\frac{\exp\left(n\log\lambda_{\max}\left(\frac{1}{n}\sum\limits_{i=1}^{n}\mathbb{E}e^{t\mathcal{X}_{i}}\right)\right)}{\left(e_{\bigodot}^{t\mathbf{b}}\right)_{j}}\right\\}$

where the last inequality holds since we bound the trace of a TPD tensor by
the eigenvalue size with $m\times p$ multiplied by the maximum eigenvalue; and
spectral mapping theorem for $\log$ and $\exp$ functions. $\hfill\Box$

##  5 Courant-Fischer Theorem under T-product Tensors and Minimum
Eigenvalue/Eigentuple

In this section, Courant-Fischer theorem for T-product tensors will be proved
and this theorem will be used to show the relationship between the maximum
eigentuple and the minimum eigentule of TPD T-product tensors. Let us recall
1.8.

See 1.8

Proof: We will just prove the first characterization of $\mathbf{d}_{k}$. The
other can be proved similarly.

First, we wish to show that $\mathbf{d}_{k}$ is achievable. As $S_{k}$ is the
space spanned by $\\{\mathbf{U}_{j}^{[k]}\\}$ for $1\leq j\leq k$ and $0\leq
l\leq p-1$. For every $\mathbf{X}\in S_{k}$, we can express $\mathbf{X}$ as

$\displaystyle\mathbf{X}$ $\displaystyle=$
$\displaystyle\sum\limits_{j=1}^{k}\sum\limits_{l=0}^{p-1}\alpha_{j}^{[l]}\mathbf{U}_{j}^{[l]}.$
(142)

Then, we have

$\displaystyle\left(\mathbf{X}^{\mathrm{H}}\star\mathcal{A}\star\mathbf{X}\right)\bigg{/}_{\bigodot}\left(\mathbf{X}^{\mathrm{H}}\star\mathbf{X}\right)$
$\displaystyle=$
$\displaystyle\sum\limits_{j=1}^{k}\sum\limits_{l=0}^{p-1}\left(\alpha_{j}^{[l]}\right)^{2}\mathbf{d}_{j}\bigg{/}_{\bigodot}\sum\limits_{j=1}^{k}\sum\limits_{l=0}^{p-1}\left(\alpha_{j}^{[l]}\right)^{2}\mathbf{e}$
(143) $\displaystyle\geq$
$\displaystyle\sum\limits_{j=1}^{k}\sum\limits_{l=0}^{p-1}\left(\alpha_{j}^{[l]}\right)^{2}\mathbf{d}_{k}\bigg{/}_{\bigodot}\sum\limits_{j=1}^{k}\sum\limits_{l=0}^{p-1}\left(\alpha_{j}^{[l]}\right)^{2}\mathbf{e}$
$\displaystyle=$ $\displaystyle\mathbf{d}_{k}$

where $\mathbf{e}=(1,0,\cdots,0)^{\mathrm{T}}\in\mathbb{C}^{p}$.

To verify that this is the maximum eigentuple, as $T_{k}$ is the space spanned
by $\\{\mathbf{U}_{j}^{[l]}\\}$ for $k\leq j\leq m$ and $0\leq l\leq p-1$, for
any $S_{k}$ with dimension $k\times p$ the intersection of $S_{k}$ with
$T_{k}$ is non-empty. Then, we also have

$\displaystyle\min\limits_{\mathbf{X}\in
S_{k}}\left(\mathbf{X}^{\mathrm{H}}\star\mathcal{A}\star\mathbf{X}\right)\bigg{/}_{\bigodot}\left(\mathbf{X}^{\mathrm{H}}\star\mathbf{X}\right)$
$\displaystyle\leq$ $\displaystyle\min\limits_{\mathbf{X}\in S_{k}\cap
T_{k}}\left(\mathbf{X}^{\mathrm{H}}\star\mathcal{A}\star\mathbf{X}\right)\bigg{/}_{\bigodot}\left(\mathbf{X}^{\mathrm{H}}\star\mathbf{X}\right).$
(144)

Any such $\mathbf{X}$ can be expressed as

$\displaystyle\mathbf{X}$ $\displaystyle=$
$\displaystyle\sum\limits_{j=k}^{m}\sum\limits_{l=0}^{p-1}\alpha_{j}^{[l]}\mathbf{U}_{j}^{[l]},$
(145)

then, we have

$\displaystyle\left(\mathbf{X}^{\mathrm{H}}\star\mathcal{A}\star\mathbf{X}\right)\bigg{/}_{\bigodot}\left(\mathbf{X}^{\mathrm{H}}\star\mathbf{X}\right)$
$\displaystyle=$
$\displaystyle\sum\limits_{j=k}^{m}\sum\limits_{l=0}^{p-1}\left(\alpha_{j}^{[l]}\right)^{2}\mathbf{d}_{j}\bigg{/}_{\bigodot}\sum\limits_{j=k}^{m}\sum\limits_{l=0}^{p-1}\left(\alpha_{j}^{[l]}\right)^{2}\mathbf{e}$
(146) $\displaystyle\leq$
$\displaystyle\sum\limits_{j=k}^{m}\sum\limits_{l=0}^{p-1}\left(\alpha_{j}^{[l]}\right)^{2}\mathbf{d}_{k}\bigg{/}_{\bigodot}\sum\limits_{j=k}^{m}\sum\limits_{l=0}^{p-1}\left(\alpha_{j}^{[l]}\right)^{2}\mathbf{e}$
$\displaystyle=$ $\displaystyle\mathbf{d}_{k}.$

Therefore, all subspace of $S_{k}$ with dimension $k\times p$, we have

$\displaystyle\min\limits_{\mathbf{X}\in
S_{k}}\left(\mathbf{X}^{\mathrm{H}}\star\mathcal{A}\star\mathbf{X}\right)\bigg{/}_{\bigodot}\left(\mathbf{X}^{\mathrm{H}}\star\mathbf{X}\right)\leq\mathbf{d}_{k}.$
(147)

This theorem is proved since $\mathbf{d}_{k}$ is achievable and is the maximum
eigentuple. $\hfill\Box$

By applying Theorem 1.8, we have following relations:

$\displaystyle\mathbf{d}_{\min}(\mathcal{X})=-\mathbf{d}_{\max}(-\mathcal{X})\leavevmode\nobreak\
\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\ \leavevmode\nobreak\
\lambda_{\min}(\mathcal{X})=-\lambda_{\max}(-\mathcal{X})$ (148)

##  6 Conclusion

In this Part I work, we try to establish following inequalities about
T-product tensors: (1) trace function nondecreasing/convexity; (2) Golden-
Thompson inequality for T-product tensors; (3) Jensen’s T-product inequality;
(4) Klein’s T-product inequality. All these inequalities are used to
generalize celebrated Lieb’s concavity theorem from matrices to T-product
tensors. Then, this new version of Lieb’s concavity theorem under T-product
tensor is utilized to build master tail bounds for the maximum eigenvalue and
the maximum eigentuple induced by independent sums of random Hermitian
T-product. In order to find the relationship between the minimum eigentuple
and the maximum eigentuple, we also extended the Courant-Fischer Theorem from
matrices to T-product tensors. How these new inequalities and Courant-Fischer
Theorem under T-product are used to derive new tail bounds of the extreme
eigenvalue and eigentuple for sums of random T-product tensors is the main
goal of our Part II paper.

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