Why the set of all $n{×}n$ PSD matrices is not a subspace of all $n{×}n$ symmetric matrices?

Question

I know this question is kind of basic, but I am pretty confused about that.

PSD is also symmetric, how can it not belongs to the symmetric's subspace? I have no idea to prove it.

Answer 1

Because the set of positive semidefinite matrices is not closed under multiplication by scalars. Suppose $A$ is a positive definite matrix, and hence positive semidefinite, and $\lambda > 0$ is an eigenvalue. If we multiply $A$ by $s < 0$, we get a negative definite matrix since:

$$Ax = \lambda x \implies (sA)x = (s \lambda) x.$$

So the problem is not that the set of positive definite matrices is not contained in the symmetric subspace, but that it is not itself a vector space.