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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.

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migrated from Oct 16 '12 at 21:46

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Use the definitions directly! The only "tool" you should need is a single application of the spectral decomposition of one of the matrices ($A$ or $B$, your choice!). – cardinal Oct 16 '12 at 20:47
Are A and B the same dimensions? – Bitwise Oct 16 '12 at 21:01
@Bitwise: They don't have to be, no. – cardinal Oct 16 '12 at 21:06
BTW I believe the new eigenvectors are the product of the pairs of A and B eigenvectors, not the sum. It still produces positive eigenvectors, though. – Bitwise Oct 16 '12 at 21:06
Use mix product property and diagonaliztion of $A$ and $B$. – Davide Giraudo Oct 16 '12 at 21:59

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.

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