Convergence from spectrum of the Choi matrix?

Question

Suppose I have a linear operation of the form $T(C)=\frac{1}{m}\sum_i^m A_i C A_i$ for a set of $m$ symmetric $d \times d$ matrices $A_i$.

I construct Choi matrix below with $e_i$ referring to standard basis

$$ M=\left(\begin{array}{cccc} T(e_1 e_1^T) & T(e_1 e_2^T )& \ldots& T(e_1 e_d^T)\\ \ldots & \ldots & \ldots&\ldots\\ T(e_d e_1^T) & T(e_d e_2^T) & \ldots & T(e_d e_d^T) \end{array} \right) $$

Is it possible to tell if $T$ is contractive or convergent by looking at eigenvalues of $M$?

Note that entries of $M$ are rearrangement of entries of $T$ viewed as linear matrix acting on $\operatorname{vec}C$ (Choi-Jamiołkowski isomorphism)

$$ T=\left(\begin{array}{c} \operatorname{vec}T(e_1 e_1^T)^T\\ \operatorname{vec}T(e_2 e_1^T)^T\\ \ldots \\ \operatorname{vec}T(e_d e_1^T)^T\\ \operatorname{vec}T(e_1 e_2^T)^T\\ \ldots \\ \operatorname{vec}T(e_d e_d^T)^T\\ \end{array} \right) $$

In my application, most eigenvalues of $M$ appear zero, while most eigenvalues of $T$ not-zero, so it seems that $M$ spectrum is easier to analyze.

Example:

$$A=\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right),\left( \begin{array}{cc} 0 & -1 \\ -1 & 0 \\ \end{array} \right) $$

$$M=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & \frac{1}{2} \\ \end{array} \right)$$

$$T=\left( \begin{array}{cccc} 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ \end{array} \right)$$

Eigenvalues of $M$ are ${1, 1/2, 0, 0}$, while eigenvalues of $T$ are $\left\{\frac{1}{4} \left(\sqrt{5}+1\right),-\frac{1}{2},\frac{1}{2},\frac{1}{4} \left(1-\sqrt{5}\right)\right\}$

Notebook

Answer 1

Since the $T$ and $M$ are basically a rearrangement of elements, their Frobenious norms are the same.

The eigenvalues satisfy

$$\sum_{i=1}^{d^2} \lambda_i(M) = tr(M) = \frac{1}{m} \sum_{i=1}^m \|A_i\|_F^2. $$

Then

$$\sum_{i=1}^{d^2} \lambda_i(M)^2 = \|M\|_{F}^2\le (\frac{1}{m}\max_{i=1}^{m} \|A_i\|_{F}^2 )^2 $$

then the trivial bound is $\rho(T) \le \sum_{i=1}^{d^2} \lambda_i(M). $ It might not be quite useful if the RHS is greater than 1.

We might need additional information for a better estimate. For $T$ (which is also symmetric), the spectral radius of $T$ is smaller than its $\|T\|_1$, which is

$$\max_{k,l} \frac{1}{m} \sum_{i=1}^m \sum_{s,t} |A_{i, k, s} A_{i, l, t}| = \max_{k,l} \frac{1}{m} \sum_{i=1}^m { \sum_{s=1}^d |A_{i, k, s}| \sum_{t=1}^d |A_{i, l, t}|} = \frac{1}{m} \sum_{i=1}^m \max_{k=1}^d\left| \sum_{s=1}^d |A_{i, k, s}| \right|^2 $$

therefore $$\rho(T)\le \frac{1}{m}\sum_{i=1}^m \max_{k=1}^d \left| \sum_{s=1}^d |A_{i, k, s}| \right|^2 = \frac{1}{m} \sum_{i=1}^m \|A_i\|_1^2$$

When the matrices $A_i$ satisfy that $ \|A_i\|_1 \le \beta \|A_i\|_F$, for some $\beta < 1$ uniformly, then

$$\rho(T) \le \beta \sum_{i=1}^{d^2} \lambda_i(M). $$