Perturbation Analysis on T-Eigenvalues of Third-Order Tensors

Abstract

This paper concentrates on perturbation theory concerning the tensor T-eigenvalues within the framework of tensor-tensor multiplication. Notably, it serves as a cornerstone for the extension of semidefinite programming into the domain of tensor fields, referred to as T-semidefinite programming. The analytical perturbation analysis delves into the sensitivity of T-eigenvalues for third-order tensors with square frontal slices, marking the first main part of this study. Three classical results from the matrix domain into the tensor domain are extended. Firstly, this paper presents the Gershgorin disc theorem for tensors, demonstrating the confinement of all T-eigenvalues within a union of Gershgorin discs. Afterward, generalizations of the Bauer-Fike theorem are provided, each applicable to different cases involving tensors, including those that are F-diagonalizable and those that are not. Lastly, the Kahan theorem is presented, addressing the perturbation of a Hermite tensor by any tensors. Additionally, the analysis establishes connections between the T-eigenvalue problem and various optimization problems. The second main part of the paper focuses on tensor pseudospectra theory, presenting four equivalent definitions to characterize tensor $\varepsilon$-pseudospectra. Accompanied by a thorough analysis of their properties and illustrative visualizations, this section also explores the application of tensor $\varepsilon$-pseudospectra in identifying more T-positive definite tensors.

Appendix

Proof of Theorem 3

The theorem is evident when $\mu \in \varLambda (\mathcal {A})$, as the left-hand sides of (12) and (13) vanish. Therefore, we assume that $\mu \notin \varLambda (\mathcal {A})$. By Lemma 1, we can see that

$$\begin{aligned} \mu I_{mn} - {\text {bcirc}}(\mathcal {A} + \varepsilon \mathcal {B}) = \mu I_{mn} - {\text {bcirc}}(\mathcal {A}) - {\text {bcirc}}(\varepsilon \mathcal {B}), \end{aligned}$$

and moreover $\mu I_{mn} - {\text {bcirc}}(\mathcal {A} + \varepsilon \mathcal {B})$ is singular. This means that

$$\begin{aligned} (F_n^{\textrm{H}} \otimes I_m) {\text {bcirc}}(\mathcal {Q})^{-1} [ \mu I_{mn} - {\text {bcirc}}(\mathcal {A}) - {\text {bcirc}}(\varepsilon \mathcal {B})] {\text {bcirc}}(\mathcal {Q}) (F_n \otimes I_m) \end{aligned}$$

(16)

is also singular since the matrices multiplied on the left and right sides are nonsingular. Notice that

$$\begin{aligned} \begin{aligned}&(F_n^{\textrm{H}} \otimes I_m) {\text {bcirc}}(\mathcal {Q})^{-1} {\text {bcirc}}(\mathcal {A}) {\text {bcirc}}(\mathcal {Q}) (F_n \otimes I_m)\\&\quad = (F_n^{\textrm{H}} \otimes I_m) [ {\text {bcirc}}(\mathcal {D}) + {\text {bcirc}}(\mathcal {N}) ] (F_n \otimes I_m)\\&\quad = \left[ \begin{array}{llll} D^{(1)} &{} &{} &{} \\ &{} D^{(2)} &{} &{} \\ &{} &{} \ddots &{} \\ &{} &{} &{} D^{(n)} \end{array}\right] + \left[ \begin{array}{llll} N^{(1)} &{} &{} &{} \\ &{} N^{(2)} &{} &{} \\ &{} &{} \ddots &{} \\ &{} &{} &{} N^{(n)} \end{array}\right] \\:=&D+N. \end{aligned} \end{aligned}$$

Therefore (16) can be rewritten as

$$\begin{aligned} \mu I_{mn} - D - N - (F_n^{\textrm{H}} \otimes I_m) {\text {bcirc}}(\mathcal {Q})^{-1} {\text {bcirc}}(\varepsilon \mathcal {B}) {\text {bcirc}}(\mathcal {Q}) (F_n \otimes I_m), \end{aligned}$$

and thus the following matrix

$$\begin{aligned} I_{mn} - (\mu I_{mn} - D - N)^{-1} (F_n^{\textrm{H}} \otimes I_m) {\text {bcirc}}(\mathcal {Q})^{-1} {\text {bcirc}}(\varepsilon \mathcal {B}) {\text {bcirc}}(\mathcal {Q}) (F_n \otimes I_m) \nonumber \\ \end{aligned}$$

(17)

is singular.

By the assumption that $|N|^q = 0$ and note that $\mu I_{mn} - D$ is a nonsingular diagonal matrix, it follows that $((\mu I_{mn} - D)^{-1}N)^q = 0$. Hence,

$$\begin{aligned} ((\mu I_{mn}-D)-N)^{-1}=\sum _{k=0}^{q-1}\left( (\mu I_{mn}-D)^{-1} N\right) ^{k}(\mu I_{mn}-D)^{-1}, \end{aligned}$$

and

$$\begin{aligned} \Vert ((\mu I_{mn}-D)-N)^{-1}\Vert \le \frac{1}{ \min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |} \sum _{k=0}^{q-1}\left( \frac{\Vert N\Vert }{ \min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |}\right) ^{k} \end{aligned}$$

holds under the 1-, 2- and $\infty$-norms cases.

If $\min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |\ge 1$, then

$$\begin{aligned} \Vert ((\mu I_{mn}-D)-N)^{-1}\Vert \le \frac{1}{ \min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |} \sum _{k=0}^{q-1}\Vert N\Vert ^{k}. \end{aligned}$$

In the case of $\min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |<1$, then

$$\begin{aligned} \Vert ((\mu I_{mn}-D)-N)^{-1}\Vert \le \frac{1}{ (\min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |)^q} \sum _{k=0}^{q-1}\Vert N\Vert ^{k}. \end{aligned}$$

By (17), we can obtain

$$\begin{aligned} \begin{aligned} 1&\le \left\| (\mu I_{mn} - D - N)^{-1} (F_n^{\textrm{H}} \otimes I_m) \cdot {\text {bcirc}}(\mathcal {Q})^{-1} {\text {bcirc}}(\varepsilon \mathcal {B}) {\text {bcirc}}(\mathcal {Q}) \cdot (F_n \otimes I_m)\right\| \\&=\left\| (\mu I_{mn} - D - N)^{-1}\right\| \Vert {\text {bcirc}}(\varepsilon \mathcal {B}) \Vert \Vert F_n^{\textrm{H}} \otimes I_m\Vert \Vert F_n \otimes I_m\Vert \Vert {\text {bcirc}}(\mathcal {Q})^{-1} \Vert \Vert {\text {bcirc}}(\mathcal {Q}) \Vert . \end{aligned} \end{aligned}$$

Combining the above theoretical analyses, in the spectral norm case, we can see

$$\begin{aligned} \min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu | \le \Vert {\text {bcirc}}(\varepsilon \mathcal {B}) \Vert _2 \sum _{k=0}^{q-1}\Vert N\Vert _2^{k} \end{aligned}$$

or

$$\begin{aligned} (\min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |)^q \le \Vert {\text {bcirc}}(\varepsilon \mathcal {B}) \Vert _2 \sum _{k=0}^{q-1}\Vert N\Vert _2^{k} \end{aligned}$$

for

$$\begin{aligned} \min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |\ge 1 \quad \text {or } \quad \min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu |<1, \end{aligned}$$

respectively. Let $\theta = \Vert {\text {bcirc}}(\varepsilon \mathcal {B}) \Vert _2 \sum _{k=0}^{q-1}\Vert N\Vert _2^{k}$. Then we get the result (12) for the spectral norm. Similar to the proof given in Theorem 2, the Frobenius norm case can be obtained easily.

For the 1- and $\infty$-norms, by using (17) again we get

$$\begin{aligned} \min _{\lambda \in \varLambda (\mathcal {A})}|\lambda -\mu | \le \max \{\theta _p, \theta _p^{1/q}\}, \end{aligned}$$

where

$$\begin{aligned} \theta _p = \Vert {\text {bcirc}}(\varepsilon \mathcal {B}) \Vert _p \kappa _{p}(\mathcal {Q}) \kappa _{p}(F_n \otimes I_m) \sum _{k=0}^{q-1}\Vert N\Vert _2^{k}. \end{aligned}$$

The proof is completed. $\square$

Proof of Theorem 4

We only need to consider the case that $\mu$ is not a T-eigenvalue of $\mathcal {A}$. Hence $\mu I_{mn} - \tilde{D}-\tilde{N}$ is nonsingular. Similar as (17), the matrix

$$\begin{aligned} I_{mn} - (\mu I_{mn}- \tilde{D} - \tilde{N})^{-1} (F_n \otimes I_m) {\text {bcirc}}(\mathcal {X})^{-1} {\text {bcirc}}(\varepsilon \mathcal {B}) {\text {bcirc}}(\mathcal {X}) (F_n^{\textrm{H}} \otimes I_m) \end{aligned}$$

is singular. By using a similar proof process given for Theorem 3 in the above, we could get the conclusion. $\square$

Proof of Theorem 8

To prove the assertion of (I), we utilize (14) and Remark 5, which allow us to exploit the properties of $\varLambda _{\varepsilon }(A^{(i)})$ for each matrix $A^{(i)}$. It is well-known that $\varLambda _{\varepsilon }(A^{(i)})$ is nonempty, open, and bounded, with at most m connected components, each containing one or more eigenvalues of $A^{(i)}$ [52, Theorem 2.4]. Consequently, these same properties also hold for the given tensor $\mathcal {A}$ by the above analysis. Additionally, the number of connected components is bounded by nm due to the relationship expressed by

$$\begin{aligned} \varLambda _{\varepsilon }(\mathcal {A})= \bigcup _{i=1}^{n} \varLambda _{\varepsilon }(A^{(i)}). \end{aligned}$$

Now, we proceed to part (II). Denote $\left( F_{n}^{\textrm{H}} \otimes I_{m}\right) {\text {bcirc}}(\mathcal {A}) \left( F_{n}\otimes I_{m}\right) :=A$. First, note that

$$\begin{aligned} {\text {bcirc}}(\mathcal {A}+c\mathcal {E})&= \left[ \begin{array}{ccccc} {A_{1} + cI_m} &{} {A_{n}} &{} {A_{n-1}} &{} {\cdots } &{} {A_{2}} \\ {A_{2}} &{} {A_{1} + c I_m } &{} {A_{n}} &{} {\cdots } &{} {A_{3}} \\ {\vdots } &{} {\ddots } &{} {\ddots } &{} {\ddots } &{} {\vdots } \\ {A_{n}} &{} {A_{n-1}} &{} {\ddots } &{} {A_{2}} &{} {A_{1} + cI_m } \end{array}\right] \\&= {\text {bcirc}}(\mathcal {A}) + c {\text {bcirc}}(\mathcal {E}) \\&= \left( F_{n} \otimes I_{m}\right) A \left( F_{m}^{\textrm{H}}\otimes I_{n}\right) + c[\left( F_{m} \otimes I_{n}\right) I_{mn} \left( F_{m}^{\textrm{H}}\otimes I_{n}\right) ]\\&= \left( F_{n} \otimes I_{m}\right) (A+cI_{mn}) \left( F_{n}^{\textrm{H}}\otimes I_{m}\right) \\&= \left( F_{n} \otimes I_{m}\right) \left[ \begin{array}{cccc} {A^{(1)}+cI_m} &{} {} &{} {} &{} {} \\ {} &{} {A^{(2)}+cI_m} &{} {} &{} {} \\ {} &{} {} &{} {\ddots } &{} {} \\ {} &{} {} &{} {} &{} {A^{(n)}+cI_m} \end{array}\right] \left( F_{n}^{\textrm{H}}\otimes I_{m}\right) . \end{aligned}$$

Therefore, for any $c\in \mathbb {C}$, we have

$$\begin{aligned} \varLambda _{\varepsilon }(\mathcal {A}+c)= \bigcup _{i=1}^{n} \varLambda _{\varepsilon }(A^{(i)}+cI_m) = \bigcup _{i=1}^{n} [\varLambda _{\varepsilon }(A^{(i)}) +c] = c + \varLambda _{\varepsilon }(\mathcal {A}). \end{aligned}$$

We complete the proof of this part.

For part (III), by Lemma 1, we know that

$$\begin{aligned} {\text {bcirc}}(c \mathcal {A})=c {\text {bcirc}}(\mathcal {A}) = c [(F_{n} \otimes I_{m}) A \left( F_{n}^{\textrm{H}}\otimes I_{m}\right) ] = \left( F_{n} \otimes I_{m}\right) (cA) \left( F_{n}^{\textrm{H}}\otimes I_{m}\right) , \end{aligned}$$

which implies that

$$\begin{aligned} \varLambda _{|c| \varepsilon }(c \mathcal {A}) = \bigcup _{i=1}^{n} \varLambda _{|c|\varepsilon }(cA_i) = \bigcup _{i=1}^{n}c \varLambda _{\varepsilon }(A_i) = c \bigcup _{i=1}^{n} \varLambda _{\varepsilon }(A_i), \end{aligned}$$

since for any nonzero $c\in \mathbb {C}$ and matrix $A\in \mathbb {C}^{m\times m}$, the following equality

$$\begin{aligned} \varLambda _{|c|\varepsilon }(cA) = c \varLambda _{\varepsilon }(A) \end{aligned}$$

holds [52, Theorem 2.4]. Thus we get the result that $\varLambda _{|c| \varepsilon }(c \mathcal {A})=c \varLambda _{\varepsilon }(\mathcal {A})$ for any nonzero $c \in \mathbb {C}$.

Now, we prove the last part of this theorem. By Lemma 1, we know that

$$\begin{aligned} {\text {bcirc}}\left( \mathcal {A}^{\textrm{H}}\right) = \left( F_{n} \otimes I_{m}\right) A^{\textrm{H}} \left( F_{n}^{\textrm{H}}\otimes I_{m}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \varLambda _{\varepsilon }\left( \mathcal {A}^{{\textrm{H}} }\right) = \bigcup _{i=1}^{n} \varLambda _{\varepsilon }((A^{(i)})^{\textrm{H}} ) = \bigcup _{i=1}^{n}\overline{ \varLambda _{\varepsilon }(A^{(i)})} = \overline{ \bigcup _{i=1}^{n}\varLambda _{\varepsilon }(A^{(i)})} = \overline{ \varLambda _{\varepsilon }\left( \mathcal {A}\right) }, \end{aligned}$$

where the conclusion $\varLambda _{\varepsilon }(A^{\textrm{H}}) = \overline{ \varLambda _{\varepsilon }(A)}$ under the two-norm case for any matrix $A\in \mathbb {C}^{m\times m}$ [52, Theorem 2.4] is applied in the second equality. $\square$

Proof of Theorem 9

If $\lambda$ is a T-eigenvalue of tensor $\mathcal {A}$, then it is an eigenvalue of the matrix ${\text {bcirc}}(\mathcal {A})$. Therefore, for any $\mu \in \mathbb {C}$, $\lambda + \mu$ is an eigenvalue of ${\text {bcirc}}(\mathcal {A}) + \mu I$. Note that $\Vert \mu I\Vert = |\mu |$, and by the definition of pseudospectra on tensors, we obtain $\lambda + \mu \in \varLambda _{\varepsilon }(\mathcal {A})$ for any $|\mu | < \varepsilon$. Thus, we have completed the proof of (15).

For the normal tensor case, by Lemmas 4 and 1, we obtain

$$\begin{aligned} {\text {bcirc}}(\mathcal {U})^{\textrm{H}} {\text {bcirc}}(\mathcal {A})({\text {bcirc}}(\mathcal {U})) = {\text {bcirc}}(\mathcal {D}), \end{aligned}$$

and

$$\begin{aligned} (F_n^{\textrm{H}} \otimes I_m) {\text {bcirc}}(\mathcal {D}) (F_n\otimes I_m) = \left[ \begin{array}{cccc} {D^{(1)}} &{} {} &{} {} &{} {} \\ {} &{} {D^{(2)}} &{} {} &{} {} \\ {} &{} {} &{} {\ddots } &{} {} \\ {} &{} {} &{} {} &{} {D^{(n)}} \end{array}\right] : = D, \end{aligned}$$

(18)

in which $D^{(i)}$ is diagonal for $i = 1, \cdots , n$ by Lemma 3. Also note that $\Vert \cdot \Vert =\Vert \cdot \Vert _{2}$, we may assume directly that $\mathcal {A}$ is F-diagonal. Therefore, the diagonal entries of ${\text {bcirc}}(\mathcal {A})$ are exactly the T-eigenvalues. As we all know, the $\varepsilon$-pseudospectra is just the union of the open $\varepsilon$-balls about the points of the spectra for any normal matrix; equivalently, we have

$$\begin{aligned} \left\| (z-{\text {bcirc}}(\mathcal {A}))^{-1}\right\| _{2}=\frac{1}{{\text {dist}}(z, \varLambda ({\text {bcirc}}(\mathcal {A})))}, \end{aligned}$$

which implies

$$\begin{aligned} {\text {dist}}(z, \varLambda ({\text {bcirc}}(\mathcal {A}))) < \varepsilon \end{aligned}$$

by the $\varepsilon$-pseudospectra of tensors. We get the conclusion since $\varLambda (\mathcal {A})+\varDelta _{\varepsilon }$ is the same as $\{z: {\text {dist}}(z, \varLambda (\mathcal {A}))<\varepsilon \}$. $\square$