3

Let $m,n\in\mathbb{N}$, with $m,n>1$. Suppose $K\in \mathbb{M}_{mn\times mn}(\mathbb{C})$ is positive semidefinite. We can always write

$$K=\sum_{i,j=1}^m E_{i,j}\otimes K_{i,j},$$

for some collection of matrices $K_{i,j}\in \mathbb{M}_{n\times n}(\mathbb{C}) $, where $E_{i,j}\in \mathbb{M}_{m\times m}(\mathbb{C})$ is the matrix with 1 in the entry $(i,j)$ and zeros everywhere else. This amounts to writing $K$ in the block diagonal form

$$\begin{pmatrix} K_{1,1} &\dots &K_{1,m} \\ \vdots &\ddots &\vdots \\ K_{m,1}& \dots & K_{m,m} \end{pmatrix}$$.

Under what conditions can we find a collection of non-square matrices $A_k\in \mathbb{M}_{mn\times n}(\mathbb{C})$ such that $K_{i,j}=A_i^*A_j$ for all $i,j\in \{1,\dots,m \}$?

In other words, when can we write

$$K=\begin{pmatrix} A_1^*A_1 &\dots &A_1^*A_m \\ \vdots &\ddots &\vdots \\ A_m^*A_1& \dots & A_m^*A_m \end{pmatrix}=\begin{pmatrix} A^*_{1}\\ \vdots\\ A^*_{m}\end{pmatrix} \begin{pmatrix} A_{1} &\dots &A_{m} \end{pmatrix}$$

for some collection of matrices $A_k\in \mathbb{M}_{mn\times n}(\mathbb{C})$?

JQX
  • 159

2 Answers2

3

You can always find those matrices $A_k$. Since $K$ is positive, there exists $B\in M_{mn}(\mathbb C)$ such that $K=B^*B$. Now, writing $B$ in the same block form as $K$, let $$ A_k=\begin{bmatrix}B_{1k}\\ B_{2k}\\ \vdots \\ B_{mk}\end{bmatrix}. $$ Then $$ K_{ij}=\sum_{h=1}^m (B^*)_{ih}B_{hj}=\sum_{h=1}^mB_{hi}^*B_{hj}=A_i^*A_j^*. $$

(this below was the answer to the original form of the question, which received upvotes and has some value, so I'm leaving it)

Note that if $K$ is of the form you want, the eigenvalues of $K$ are the same (bar zeroes) to those of $$ \begin{bmatrix}A_1&\cdots&A_m\end{bmatrix}\begin{bmatrix}A_1^*\\ \vdots \\A_m^*\end{bmatrix} =A_1A_1^*+\cdots+A_mA_m^*, $$ an $mn\times mn$ matrix. So $K$ has at most $mn$ distinct eigenvalues, which discards many possible $K$.

What happens in general is that $K$ is a sum of matrices of that form. Since $K$ is positive semidefinite, there exists $B$, with blocks the same size as those of $K$, such that $K=BB^*$. Now let $$ B_k= \begin{bmatrix} 0&\cdots&0&B_{k1}&0\cdots&0\\ 0&\cdots&0&B_{k2}&0\cdots&0\\ 0&\cdots&0&\vdots&0\cdots&0\\ 0&\cdots&0&B_{km}&0\cdots&0\\ \end{bmatrix}. $$ Then, as $B_jB_k^*=0$ if $j\ne k$, $$ K=BB^*=\sum_{k=1}^m B_k\sum_{k=1}^mB_k^*=\sum_{k=1}^mB_kB_k^* =\sum_{k=1}^m\begin{bmatrix}B_{k1}\\ \vdots\\ B_{km}\end{bmatrix} \begin{bmatrix}B_{k1}^*&\cdots& B_{km}^*\end{bmatrix}. $$ Now you can take $A_{kj}=B_{kj}^*$.

Martin Argerami
  • 205,756
  • Thanks Martin. I realize that I made a typo: I wrote $A_k\in \mathbb{M}{n\times mn}(\mathbb{C})$ which doesn't make sense, and should instead be $A_k\in \mathbb{M}{mn\times n}(\mathbb{C})$ (I've already fixed it in the question). As a result, the matrix in the first line of your answer is actually an $mn\times mn$ matrix. How would that change the conclusion? – JQX Dec 09 '15 at 15:53
  • The conclusion changes completely: the $A_k$ always exists. Please see the edit. – Martin Argerami Dec 09 '15 at 16:54
  • Thanks Martin. I reached the same conclusion as soon as I fixed my question. Sorry for the confusion. – JQX Dec 09 '15 at 16:55
  • No problem :) $ $ – Martin Argerami Dec 09 '15 at 17:14
  • One last question: in your answer to the original form of the question (which is way more interesting), you started by using the fact that for any two matrices $A$ and $B$, the nonzero points of the spectra of $AB$ and $BA$ coincide. Do you know if anything can be said about the multiplicity of these non-zero eigenvalues for both $AB$ and $BA$? i.e. if $\lambda$ is a non-zero eigenvalue of $AB$ with multiplicity $k$, can we say anything about its multiplicity as an eigenvalue of $BA$? – JQX Dec 09 '15 at 17:39
  • The multiplicity is the same. One can show that if the size of $BA$ is $n$ and the size of $AB=n+m$, then $$p_{AB}(t)=t^mp_{BA}(t).$$ – Martin Argerami Dec 09 '15 at 20:53
1

Actually the answer turns out to be trivial. Since $K$ is positive semidefinite, we can write $K=B^*B$ for some $B\in \mathbb{M}_{mn\times mn}(\mathbb{C})$. Write $$B=\begin{pmatrix} B_{11} & \dots &B_{1m} \\ \vdots& \ddots &\vdots \\ B_{m1}&\dots &B_{mm} \end{pmatrix},$$ and let $$A_{k}=\begin{pmatrix} B_{1k}\\ \vdots\\ B_{mk} \end{pmatrix}\in \mathbb{M}_{mn\times n}(\mathbb{C}).$$ Then $$B=\begin{pmatrix} A_1 &\dots &A_m \end{pmatrix}\quad\quad B^*=\begin{pmatrix} A_1^*\\\vdots \\A_m^* \end{pmatrix},$$ and $$K=B^*B=\begin{pmatrix} A_1^*\\\vdots \\A_m^* \end{pmatrix}\begin{pmatrix} A_1 &\dots &A_m \end{pmatrix}$$

JQX
  • 159