# An inequality for the spectral radius of block matrices

Let $$d,m$$ be positive integers. Suppose that $$A_{i,j}$$ is a $$d\times d$$-matrix with real entries whenever $$i,j\in\{1,\dots m\}$$.

Let $$A$$ be the $$dm\times dm$$ matrix that can be written as a block matrix as $$A=\begin{bmatrix} A_{1,1} & \cdots & A_{1,m} \\ \vdots & \ddots & \vdots \\ A_{m,1} & \dots & A_{m,m} \end{bmatrix}.$$

1. Is $$\rho(A)^2\leq\min(d,m)\cdot \rho\left(\sum_{i,j}A_{i,j}\otimes A_{i,j}\right)?$$

2. For all $$d,m$$, can we select real matrices $$A_{i,j}$$ so that $$\rho(A)^2=\min(d,m)\cdot \rho\left(\sum_{i,j}A_{i,j}\otimes A_{i,j}\right)>0?$$

3. If $$\rho(A)^2=\min(d,m)\cdot \rho\left(\sum_{i,j}A_{i,j}\otimes A_{i,j}\right)>0,$$ and $$d\leq m$$ then is $$\text{Rank}(A_{i,j})\leq 1$$ for each $$i,j$$?

My computer calculations suggest that the answer to these questions is 'yes'. This is a follow up of this question. This question can be generalized to the case when each $$A_{i,j}$$ is a complex matrix. One can rephrase the complex version of this question in terms of the completely depolarizing channel in quantum information theory.

• After I answered, I noticed that you asked one more question at mathoverflow.net/questions/422925 . It can be easily done with exactly the same idea but I am too sleepy to type anything now (it is 30 minutes past midnight here) so, I hope, you'll figure it out by yourself. :-) May 29, 2022 at 4:40

I'll answer #1 only, leaving the rest to you or someone else to figure out. The answer is affirmative.

Dropping the trivial case $$m=1$$, we may assume by density that we are in the generic position, i.e., that $$A_{ij}^T\ne 0$$ and together have no non-trivial invariant subspace (this is just to avoid considering degenerate cases, which can also be done using dimension reduction but I prefer not to bother with this extra technicality).

Consider the mapping $$X\mapsto F(X)=\sum_{i,j}A_{ij}XA_{ij}^T$$ on the cone $$K$$ of symmetric non-negative definite $$d\times d$$ matrices $$X$$. Notice that if $$X\ne 0$$ and $$F(X)=0$$, then all $$A_{ij}^T$$ map the entire space to the kernel of $$X$$, which contradicts our generic position assumption. Thus the mapping $$X\mapsto \frac 1{\operatorname{Tr}F(X)}F(X)$$ is well-defined on the convex compact $$Q=K\cap\{X:\operatorname{Tr}X=1\}$$ and maps $$Q$$ to itself. So, by the Brouwer fixed point theorem, the linear mapping $$F$$ has an eigenvector in the cone with positive eigenvalue, which we can normalize to $$1$$ by scaling $$A$$, i.e., we have $$\sum_{i,j}A_{ij}XA_{ij}^T=X\,,$$ which, clearly, implies $$\rho(\sum_{i,j}A_{ij}\otimes A_{ij})\ge 1$$.

I now claim that $$X$$ is non-degenerate. Indeed, otherwise the kernel of $$X$$ would be non-trivial and invariant under all $$A_{ij}^T$$.

Now for a vector $$y=(y_1,\dots,y_m)\in\mathbb R^{md}$$ define its norm by $$\|y\|^2=\sum_j|X^{-1/2}y_j|^2$$ where $$|\cdot|$$ stands for the usual Euclidean norm in $$\mathbb R^d$$. Let $$y$$ be a unit vector (in this norm) on which the induced operator norm of $$A$$ is attained.

Then $$\|Ay\|^2=\sum_{i} \langle X^{-1/2}(Ay)_i,\xi_i\rangle^2$$ for some unit vectors $$\xi_i\in\mathbb R^d$$. Now use Cauchy-Schwarz: $$\langle X^{-1/2}(Ay)_i,\xi_i\rangle^2= \left[\sum_j \langle X^{-1/2}A_{ij}y_j,\xi_i\rangle\right]^2 \\ = \left[\sum_j \langle X^{-1/2}y_j,X^{1/2}A_{ij}^TX^{-1/2}\xi_i\rangle\right]^2 \\ \le \left[\sum_j |X^{-1/2}y_j|^2\right]\left[\sum_j |X^{1/2}A_{ij}^TX^{-1/2}\xi_i|^2\right]$$ This allows one to estimate the induced operator norm of $$A$$ and, thereby $$\rho(A)$$ by $$\rho(A)^2\le\|A\|^2\le \sum_{i,j}|X^{1/2}A_{ij}^TX^{-1/2}\xi_i|^2 \\ =\sum_{i,j}\langle X^{-1/2}A_{ij}XA_{ij}^TX^{-1/2}\xi_i,\xi_i\rangle\,.$$ Now let $$L$$ be the linear subspace of $$\mathbb R^d$$ spanned by all vectors $$\xi_i$$. Obviously, $$\operatorname{dim}L\le\min(d,m)$$. Let $$P$$ be the orthogonal projection to $$L$$. Then for every non-negative definite operator $$S$$ in $$\mathbb R^d$$, we have $$\langle S\xi_i,\xi_i\rangle\le\operatorname{Tr} (PSP)$$, which yields the bound $$\sum_{i,j} \operatorname{Tr}(PX^{-1/2}A_{ij}XA_{ij}^TX^{-1/2}P) = \operatorname{Tr}\left(\sum_{i,j} PX^{-1/2}A_{ij}XA_{ij}^TX^{-1/2}P\right) \\ = \operatorname{Tr}\left(PX^{-1/2}\left[\sum_{i,j} A_{ij}XA_{ij}^T\right]X^{-1/2}P\right)= \operatorname{Tr}(PX^{-1/2}XX^{-1/2}P) \\ =\operatorname{Tr} P=\operatorname{dim}L\le\min(d,m)$$ as required.

The answer to all three questions is Yes.

#2 This part is easy. If $$i>d$$ or $$j>d$$, then set $$A_{i,j}=0_{d\times d}$$, and if $$i\leq d$$ and $$j\leq d$$, then let $$A_{i,j}$$ be the $$d\times d$$-matrix where the $$i,j$$-th entry is $$1$$ but every other entry is zero. Then $$\rho(A)^2=\min(m,d)\cdot\rho(\sum_{i,j}A_{i,j}\otimes A_{i,j})$$. Therefore, this inequality is sharp.

#1 and #3

For generality and to put this answer in the context of quantum information theory, let me answer in the case when each $$A_{i,j}$$ is a complex matrix (we have to modify the statements of 1,2,3 though). We will only answer this question in the case when $$d\leq m$$ since fedja's answer already answers #1 for all real cases.

For this answer, we shall hold to the convention that if $$A$$ is a $$dm\times dm$$-matrix, then $$(A_{i,j})_{i,j}$$ are the submatrices of $$A$$ such that $$A$$ becomes a block matrix with blocks $$A_{i,j}\in M_{d}(\mathbb{C})$$ whenever $$1\leq i\leq m,1\leq j\leq m$$.

Let $$L(\mathbb{C}^d)$$ denote the collection of all linear operators $$A:\mathbb{C}^d\rightarrow\mathbb{C}^d$$. Suppose that $$\mathcal{E}:L(\mathbb{C}^d)\rightarrow L(\mathbb{C}^d)$$ is a linear operator. Then $$\mathcal{E}$$ is said to be trace preserving if $$\text{Tr}(\mathcal{E}(A))=\text{Tr}(A).$$ We say that $$\mathcal{E}$$ is positive if $$\mathcal{E}(A)$$ is positive semidefinite whenever $$A$$ is positive semidefinite. We say that $$\mathcal{E}$$ is completely positive if whenever $$V$$ is a finite dimensional complex Hilbert space, we have $$\mathcal{E}\otimes 1_{L(V)}:L(\mathbb{C}^{d}\otimes V)\rightarrow L(\mathbb{C}^{d}\otimes V)$$. We say that $$\mathcal{E}$$ is a quantum channel if $$\mathcal{E}$$ is completely positive and trace preserving.

The mapping $$\mathcal{E}$$ is completely positive if and only if there are $$A_1,\dots,A_r$$ with $$\mathcal{E}(X)=A_1XA_1^*+\dots+A_rXA_r^*$$ for each $$X\in L(\mathbb{C}^d)$$. The mapping $$\mathcal{E}$$ is a quantum channel if and only if there are $$A_1,\dots,A_r$$ with $$\mathcal{E}(X)=A_1XA_1^*+\dots+A_rXA_r^*$$ for each $$X\in L(\mathbb{C}^d)$$ and $$A_1^*A_1+\dots+A_r^*A_r=1_{d}$$. Observe that if $$\mathcal{E}$$ is a quantum channel, then $$\rho(\mathcal{E})=1$$.

If $$A_{i,j}$$ is an $$d\times d$$-matrix whenever $$1\leq i\leq m$$ and $$1\leq j\leq m$$, then define a superoperator $$\Phi((A_{i,j})_{i,j}):L(\mathbb{C}^d)\rightarrow L(\mathbb{C}^d)$$ by letting $$\Phi((A_{i,j})_{i,j})(X)=\sum_{i,j}A_{i,j}XA_{i,j}^*$$. When we generalize #1 to complex matrices, this statement becomes $$\rho(A)^2\leq\min(d,m)\rho(\Phi((A_{i,j})_{i,j})$$. #3 when generalized states that if $$\rho(A)^2=d\cdot\rho(\Phi((A_{i,j})_{i,j})$$ and $$d\leq m$$, then $$\text{Rank}(A_{i,j})=1$$ for each $$i,j$$.

Let $$O$$ be the collection of all systems $$(A_{i,j})_{i,j}\in M_d(\mathbb{C})^{m\times m}$$ where $$\Phi((A_{i,j})_{i,j})$$ is not nilpotent. Let $$E$$ be the collection of all systems $$(A_{i,j})_{i,j}\in M_d(\mathbb{C})^{m\times m}$$ where there is some $$\lambda>0$$ and $$B$$ where $$\Phi((\lambda BA_{i,j}B^{-1})_{i,j})$$ is a quantum channel. By the arguments that I gave in this answer, the set $$E$$ is dense in $$L(\mathbb{C}^{d})^{m\times m}$$. Let $$E^{\uparrow}$$ (respectively $$O^{\uparrow}$$) be the collection of all matrices $$A\in M_{md}(\mathbb{C})$$ where $$(A_{i,j})_{i,j}\in E$$ (respectively $$(A_{i,j})_{i,j}\in O$$).

Suppose that $$\Phi((A_{i,j})_{i,j})$$ is a quantum channel. Then

$$d=\text{Tr}(1_d)=\text{Tr}(\sum_{i,j}A_{i,j}^{*}A_{i,j})=\sum_{i,j}\|A_{i,j}\|_{2}^{2}.$$

Therefore, we have $$\rho(A)^2\leq\|A\|_{\infty}^2\leq \|A\|_{2}^2\leq d$$. Furthermore, if $$\rho(A)^2=d$$, then $$\|A\|_{\infty}=\|A\|_{2}$$, and this is only possible if $$\text{Rank}(A)=1$$ which implies that $$\text{Rank}(A_{i,j})\leq 1$$ as well.

Suppose that $$(A_{i,j})_{i,j}\in E$$. Then there is some $$\lambda>0$$ and some invertible $$B$$ where $$\Phi((\lambda BA_{i,j}B^{-1})_{i,j})$$ is a quantum channel. Suppose now that $$C$$ is the block matrix with blocks $$(\lambda BA_{i,j}B^{-1})_{i,j}$$. Then

$$\frac{\rho(A)^2}{\rho(\Phi((A_{i,j})_{i,j})}=\frac{\rho(C)^2}{\rho(\Phi((\lambda BA_{i,j}B^{-1})_{i,j})}=\rho(C)\leq d.$$

Therefore, $$\rho(A)^2\leq d\cdot \rho(\Phi((A_{i,j})_{i,j}).$$

Furthermore, if $$(A_{i,j})_{i,j}\in E$$, and $$\rho(A)^2=d\cdot \rho(\Phi((A_{i,j})_{i,j})$$, then we know that $$\text{Rank}(C)=1$$, so $$\text{Rank}(A)=1$$ as well, and therefore $$\text{Rank}(A_{i,j})=1$$ for each $$i,j$$.

Since $$E$$ is dense $$(L(\mathbb{C}^{d}))^{m\times m}$$, we conclude that $$\rho(A)^2\leq d\cdot \rho(\Phi((A_{i,j})_{i,j})$$ whenever $$(A_{i,j})_{i,j}\in (L(\mathbb{C}^{d}))^{m\times m}.$$

I now claim that whenever $$A\in L(\mathbb{C}^{m\times d})$$, if $$\rho(A)^2=d\cdot \rho(\Phi((A_{i,j})_{i,j})\neq 0$$, then $$\text{Rank}(A)=1$$.

We shall now define functions $$f:O^{\uparrow}\rightarrow\mathbb{R},g:M_{md}(\mathbb{C})\rightarrow\mathbb{R}$$. Set $$f(A)=\frac{\rho(A)^2}{\rho(\Phi((A_{i,j})_{i,j})}$$ for $$A\in O^{\uparrow}$$. Then the function $$f$$ is continuous. Let $$g(A)=\frac{|\lambda_{2}^2|}{\rho(\Phi((A_{i,j})_{i,j}))}$$ where $$\lambda_{1},\dots,\lambda_{m\cdot d}$$ are the eigenvalues of $$A$$ ordered so that $$|\lambda_{1}|\geq|\lambda_{2}|\geq\dots\geq|\lambda_{m\cdot d}|$$. The motivation behind the function $$g$$ is that for $$A\in O^{\uparrow}$$, $$\text{Rank}(A)\leq 1$$ precisely when $$g(A)=0$$ and that $$g(A)=g(C)$$ whenever there is some $$\lambda\neq 0$$ and $$B$$ where $$A_{i,j}=\lambda\cdot BC_{i,j}B^{-1}$$ for all $$i,j$$.

Let $$f(A)=d\leq m$$. Let $$\delta>0$$. Then there is some $$\epsilon>0$$ where if $$\|A-H\|_2<\epsilon$$, then $$f(H)>d-\delta$$.

Suppose that $$\epsilon>0$$. Then there is some $$H\in E^{\uparrow}$$ where $$\|A-H\|_2<\epsilon$$, and therefore $$f(H)>d-\delta$$. Now, since $$H\in E^{\uparrow}$$, there are $$\lambda,B$$ where if $$C_{i,j}=\lambda\cdot B\cdot H_{i,j}\cdot B^{-1}$$ for each $$i,j$$, then $$\Phi((C_{i,j})_{i,j})$$ is a quantum channel. Therefore, $$f(C)=f(H)>d-\delta$$.

Therefore, we have $$d-\delta\leq\rho(C)^2\leq\|C\|^2_\infty\leq\|C\|_2^2\leq d.$$ Now, let $$\lambda_1,\dots,\lambda_{md}$$ be the eigenvalues of $$C$$, and let $$\sigma_1,\dots,\sigma_{md}$$ be the singular values of $$C$$. Suppose that $$|\lambda_1|\geq|\lambda_2|\geq\dots\geq|\lambda_{md}|$$. Then $$d-\delta\leq\max(|\lambda_1|^2,\dots,|\lambda_{md}|^2)\leq |\lambda_1|^2+\dots+|\lambda_{md}|^2\leq \sigma_1^2+\dots+\sigma_{md}^2\leq d.$$

Thus $$d-\delta\leq|\lambda_1|^2\leq|\lambda_1|^2+|\lambda_2|^2\leq d$$, so $$d-\delta+|\lambda_2^2|\leq d$$. Therefore, $$|\lambda_2^2|\leq\delta$$.

We conclude that $$g(H)=g(C)\leq\delta$$.

Therefore, $$g(H)\rightarrow 0$$ as $$H\rightarrow A,H\in E^{\uparrow}$$. Therefore, $$g(A)=0$$ which means that $$\text{Rank}(A)\leq 1$$ and therefore $$\text{Rank}(A_{i,j})\leq 1$$ for all $$i,j$$.