Postitive definiteness of the Kronecker product of two positive definite matrices

Question

Let $A$ and $B$ both be positive definite matrices. How do I show that their Kronecker product is also positive definite?

I know we can use the fact that the eigenvalues of the Kronecker product is $\lambda_A+\lambda_B$ which are all positive. But I want to use a different approach here.

Thank you.

Answer 1

First approach: If $\{\lambda_1,\dots,\lambda_m\}$ are the eigenvalues of $A$ and $\{\nu_1,\dots,\nu_n\}$ those of $B$, then the eigenvalues of $A\otimes B$ are $\lambda_j\cdot\mu_k,1\leq j\leq m,1\leq k\leq n$.

We assume that the respective dimensions of $A$ and $B$ are $m$ and $n$. If $v$ is an eigenvector of $A$ for $\lambda_k$ and $w$ of $B$ for $\mu_j$, consider $V$ the vector of size $mn$, defined by $$V=(v_1w_1,\dots,v_1w_n,v_2w_1,\dots,v_2w_n,\dots,v_mw_1,v_mw_n).$$ It's an eigenvector of $A\otimes B$ for the eigenvalue $\lambda_k\mu_j$. As the matrices $A$ and $B$ are diagonalizable, counting multiplicity we are sure there aren't other eigenvalues.

As $A$ and $B$ are positive definite, $\lambda_k\mu_j>0$ for all $k,j$.

Second approach: We use mix product property, that is $$(A_1A_2)\otimes (B_1B_2)=(A_1\otimes B_1)(A_2\otimes B_2).$$ Applied twice, this gives $$A\otimes B=(P_1^tD_1P_1)\otimes (P_2^tD_2P_2),$$ where $P_i$ are orthogonal and $D_i$ diagonal. This gives $$A\otimes B=(P_1\otimes P_2)^t(D_1\otimes D_2)(P_1\otimes P_2),$$ so the problem reduces to the case $A$ and $B$ diagonal, which is easy, as the eigenvalues are positive.

Answer 2

Here is another approach: Let $A,B\geq0$, so there exists positive square roots $\sqrt{A}, \sqrt{B}$. Thus, we have (essentially by definition of the tensor/kronecker product of operators/matrices):

$$ A \otimes B = \sqrt{A} \otimes \sqrt{B} \cdot \sqrt{A} \otimes \sqrt{B}$$

But $\sqrt{A} \otimes \sqrt{B}$ is a self adjoint matrix, so it's square must be positive.

Answer 3

Let $X = (X_{i})$ and $Y = (Y_{j})$ be independent random vectors such that $X \sim \mathcal{N}(0,A)$ and $Y \sim \mathcal{N}(0,B)$. Consider $Z = (X_{i}Y_{j})$ to be a vector. It can be seen that the covariance matrix of $Z$ is exactly $A \otimes B$.