Let $A, B \in \mathbb{R}^{n \times n}$ be symmetric and positive definite matrices. Define: $$C = \begin{pmatrix} A_{11}B & A_{12}B & \cdots & A_{1n}B\\ \vdots & \vdots & \ddots & \vdots \\ A_{n1}B & A_{n2}B & \cdots & A_{nn}B \end{pmatrix} \in \mathbb{R}^{n^2 \times n^2}$$

Show that $C$ is symmetric and positive definite.

The proof that $C$ is symmetric is simple since it is a direct consequence of $A^T = A$ and $B^T = B$. The problem is to prove that $C$ is positive definite.

Let $x \in \mathbb{R}^{n^2}\setminus \{0\}$ with $x = (x_1,\dots,x_n)^T$ and $x_i \in \mathbb{R}^n$. I'd like to reduce the problem of $x^TCx > 0$ so that I can use the assumptions. Computing the regular formula for $x^TCx$ ended in a chaos for me, is there any simpler way to prove this?


1 Answer 1


This is called the Kronecker product and is denoted by $A\otimes B$.

One can show that the eigenvalues of $C$ are given by:

$$\{\lambda_C\}=\{\lambda_A\}\{\lambda_B\}$$ That is, eigenvaues of $C$ are all products of eigenvalues of $A$ and $B$.

So for $\{\lambda_A\},\{\lambda_B\}>0$, we get $\{\lambda_C\}>0$.

To prove:

$$\begin{align} (A\otimes B)(x\otimes y)&=Ax\otimes By\\ &=\lambda_Ax\otimes\lambda_By\\ &=\lambda_A\lambda_B(x\otimes y) \end{align}$$

It is also not difficult to show that $$(A\otimes B)^T=A^T \otimes B^T$$ which can be used to conclude that the Kronecker product is a symmetric matrix when $A$ and $B$ are symmetric.

  • $\begingroup$ Thank you, especially for the proof with the eigenvalues. I just checked the definition of the Kronecker product for matrices vectors $(x \otimes y) = (x_1y,x_2y,\dots,x_ny) \in \mathbb{R}^{n^2}$ and then this proof is a lot easier than my computation with 4 irritating sums. $\endgroup$
    – PeterMcCoy
    Commented Dec 10, 2016 at 22:47
  • $\begingroup$ You're welcome! $\endgroup$
    – msm
    Commented Dec 10, 2016 at 23:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .