Kronecker Product of Normal Matrices

Question

Theorem. If $A\in M_m$ and $B\in M_n$ are both normal, so is $A\otimes B$. The converse is true if $A\otimes B\ne 0$.

Proof. Suppose that $A\in M_m$ and $B\in M_n$ are both normal. Then \begin{align*} (A\otimes B)^*(A\otimes B)=(A^*\otimes B^*)(A\otimes B)&=(A^*A)\otimes(B^*B) \\ &=(AA^*)\otimes(BB^*) \\ &=(A\otimes B)(A^*\otimes B^*)=(A\otimes B)(A\otimes B)^*. \end{align*} Therefore, $A\otimes B$ is normal.

Now suppose that $A\otimes B$ is nonzero and normal. Then both $A$ and $B$ are nonzero. Then we have \begin{equation*} (A^*A)\otimes(B^*B)=(AA^*)\otimes(BB^*). \end{equation*} Consider the $i$-th diagonal block of these two products: \begin{equation*} (A^*A)_{ii}(B^*B)=\left(\sum_{j=1}^{n}{|a_{ji}|^2}\right)(B^*B)=\left(\sum_{j=1}^{n}{|a_{ij}|^2}\right)(BB^*)=(AA^*)_{ii}(BB^*). \end{equation*} Summing over all $i$'s, we can get \begin{equation*} \sum_{i=1}^{n}{\left(\sum_{j=1}^{n}{|a_{ji}|^2}\right)(B^*B)}=\sum_{i=1}^{n}{\left(\sum_{j=1}^{n}{|a_{ij}|^2}\right)(BB^*)}, \end{equation*} That is, $$\operatorname{tr}(A^*A)\cdot(B^*B)=\operatorname{tr}(AA^*)\cdot(BB^*).$$ Since $A$ is nonzero, $\operatorname{tr}(A^*A)=\operatorname{tr}(AA^*)\ne 0$, which implies that $B^*B=BB^*$ and hence $A^*A=AA^*$. $\blacksquare$

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Can anyone please help to check if my proof is valid? For the converse part, I am not confident whether such method indeed works. Besides, if you have any alternative elegant proofs for this theorem, I would be extraordinarily happy to hear from you!

Answer 1

Note: in what follows, eigenvalues and singular values are always assumed to be sorted by modulus, i.e. $\sigma_k\geq \sigma_{k+1}$ and $\big\vert\lambda_k\big\vert\geq \big\vert \lambda_{k+1}\big \vert$

The key fact is that that
$\sum_{k=1}^m \sigma_k^{(C)}\gt \sum_{k=1}^m \vert\lambda_k^{(C)}\vert $
if $C$ is not normal. (Conversely, of course, if $C$ is normal we have $\sigma_k^{(C)} =\vert\lambda_k^{(C)}\vert $ for all $k$.)

Using the above relationship we compare the singular values with (the modulus of) the eigenvalues of $A\otimes B$ noting that neither matrix is zero so e.g. their nuclear norms (Schatten 1-norms) are non-zero
$\sum_{k=1}^{m\cdot n}\sigma_{A\otimes B}$
$= \Big(\sum_{k=1}^m \sigma_k^{(A)}\big)\big(\sum_{j=1}^n \sigma_{j}^{(B)}\Big)$
$\geq \Big(\sum_{k=1}^m \big\vert \lambda_k^{(A)}\big\vert\Big)\Big(\sum_{j=1}^n \big\vert \lambda_{j}^{(B)}\big\vert\Big)$
$= \sum_{k=1}^{m\cdot n}\big\vert\lambda_{A\otimes B}\big\vert = \sum_{k=1}^{m\cdot n}\sigma_{A\otimes B}\gt 0$

where the final equality follows because $A\otimes B$ is normal. The inequality is met with equality, thus $A$ is normal and $B$ is normal.

deferred proof of the singular value inequality
Consider $C\in \mathbb C^{m\times m}$ where $\text{rank } C =r$. Using Schur Triangularization we can map $C$ to an upper triangular matrix $T$. Unitary multiplications don't change singular values so $T$ has the same singular values as $C$. Further consider $T':=DT$ where $D$ is a unitary diagonal matrix chosen so that $T'$ has all real non-negative values on its diagonal. Now run (a choice of) Polar Decomposition $T'=UP$.

We already know (i.) $\sum_{k=1}^m \vert \lambda_k^{(C)}\vert=\text{trace}\big(T'\big) $ and (ii.) $\text{C is normal }\implies \text{trace}\big(T'\big)= \sum_{k=1}^m \sigma_k$

it remains to prove: $\text{trace}\big(T'\big) =\sum_{k=1}^m \sigma_k\implies \text{C is normal }$
We do so by showing $UP =PU$; this means $(T')^*T' = P^2 =UP^2U^{-1} =T'(T')^*$ i.e. $T'$ is normal, and since it is already upper triangular this means $T'$ is diagonal (recall e.g. Schur's Inequality) hence $T$ is diagonal which is unitarily similar to $C$ so $C$ is normal as required. $P$ is Hermitian PSD so $P=V\Sigma V^{-1} = V\begin{bmatrix} \ \Sigma_{r} &\mathbf 0 \\ \mathbf 0& \mathbf 0_{m-r} \end{bmatrix}V^{*}$

$\text{trace}\big(T'\big)= \text{trace}\big(UP\big)= \text{trace}\big(U(V\Sigma V^*)\big)= \text{trace}\big((V^*UV)\Sigma\big)$
$= \text{trace}\big(Q\Sigma\big)=\sum_{k=1}^r q_{k,k}\cdot \sigma_k=\big\vert\sum_{k=1}^r q_{k,k}\cdot \sigma_k\big \vert$
$\leq \sum_{k=1}^r \big\vert q_{k,k}\big \vert \cdot \sigma_k \color{blue}{\text{ (triangle inequality)}} $
$\leq \sum_{k=1}^r \sigma_k \color{blue}{\text{ (Q is unitary)}}$
$=\sum_{k=1}^m \sigma_k \color{blue}{\text{ (rank C = r)}} $
$=\text{trace}\big(T'\big) \color{blue}{\text{ (by assumption)}}$

Since the two inequalities are met with equality we can infer for $k\in \big\{1,2,\dots , r\big\}$ that $q_{k,k} = 1 \implies q_{k,i}=0=q_{i,k}$ for $i\neq k$ as each column and row of a unitary matrix has a $2-$norm of $1$
$\implies Q\Sigma = \begin{bmatrix} \ I_{r} &\mathbf 0 \\ \mathbf 0& Q_{m-r} \end{bmatrix}\begin{bmatrix} \ \Sigma_{r} &\mathbf 0 \\ \mathbf 0& \mathbf 0_{m-r} \end{bmatrix}= \begin{bmatrix} \ \Sigma_{r} &\mathbf 0 \\ \mathbf 0& \mathbf 0_{m-r} \end{bmatrix}\begin{bmatrix} \ I_{r} &\mathbf 0 \\ \mathbf 0& Q_{m-r} \end{bmatrix}=\Sigma Q$

$\implies UP =PU$ since $U = VQV^{-1} $ and $P= V\Sigma V^{-1} $ and conjugation doesn't change commutativity.
Thus the above shows that $ \sum_{k=1}^m\vert \lambda_k^{(C)}\vert \leq \sum_{k=1}^m \sigma_k$ with equality iff $C$ is normal.