# Prove that multiplication with positive semidefinite matrix is a kernel

Let $$A \in \mathbb{R}^{d \times d}$$ be a symmetric and positive semi-definite matrix. Prove that

$$k(\textbf{x}, \textbf{x}') = \textbf{x}^T A \textbf{x}'$$

is a kernel.

My first thought when I saw this question was that $$A$$ can be interpreted to be the kernel matrix of $$k$$ and that's it. But $$A$$ and the kernel matrix of $$k$$ can be completely different, making this idea not work.

This can be proven either by showing that the kernel matrix of $$k$$ is positive semidefinite or by utilizing a kernel feature map $$\phi$$. I assume that the first approach makes more sense here, but I don't see how this can be proven just yet.

How can it be proven that $$k$$ is a kernel?

Since $$A$$ is symmetric and PSD, it can be written as $$A=Q\Lambda Q^T=(Q\Lambda^{1/2})(Q\Lambda^{1/2})^T=M^TM$$ where $$M=(Q\Lambda^{1/2})^T$$. So, the kernel can be expressed as $$k(x,x')=x^TM^TMx'=(Mx)^T(Mx')=$$ So, the corresponding transformation is $$\phi(x)=Mx$$, and $$k$$ is a kernel.
• That was very nice but you have a typo ( $Xx ^{\prime}$ ) when you write it as a dot product. Jul 24, 2021 at 17:14