Postitive definiteness of the Kronecker product of two positive definite matrices

Solution 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.

Solution 2:

The post of Davide Giraudo presupposes that one can diagonalise the matrix. This is not necessary the case even if the matrix is positive-definite. As an example, consider the matrix

$$\begin{pmatrix}1 & 1\\\ 0 & 1\end{pmatrix}$$

which is positive-definite but not diagonalizable (since it is a Jordan matrix). Of course one still can use a decomposition with generalised eigenvalues to prove that $A\otimes B$ is positive definite in case $A$ and $B$ are, but here I give a direct proof.

Let $x\neq 0$, $x\in\mathbb{R}^{nm}$, $A\in \mathbb{R}^{n\times n}$, $B\in \mathbb{R}^{m\times m}$, and both $A$ and $B$ are positive-definite. It is always possible to split

$$ x = \begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix},~x_i\in\mathbb{R}^{m}.$$

This allows us to rewrite a quadratic form $$x^{T} \left(A\otimes B\right) x = \sum_{i=1}^{n}\sum_{j=1}^{n} A_{ij} x^{T}_i B x_{j}.$$

Consider the matrix $G_{ij} = x^{T}_i B x_{j}$. It is symmetric (trivial) and positive-definite. Indeed, for $z \neq 0$ we have

$$\sum_{ij}z_{i}G_{ij} z_{j} = \left(\sum_{i} z_ix_i\right)^{T} B \left(\sum_{j} z_j x_j\right) \underset{B \text{ is positive definite}}{>} 0.$$

(in fact it is enough to say that $G$ is a Gram matrix)

Now we know that $x^{T} \left(A\otimes B\right) x = \text{tr} \left(AG\right)$, where $G$ is symmetric and positive-definite. Spectral theorem for $G$ does the rest

$$G = \sum_{k} \lambda_k v_k v_k^T \Longrightarrow \text{tr} \left(AG\right) = \sum_{k} \lambda_k v_k^TA v_k \underset{\text{as a series with positive terms}}{>} 0$$

Solution 3:

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.