[Math] If a matrix $A$, not necessarily symmetric, has real, nonnegative eigenvalues, is it positive semidefinite

Question

We know that a symmetric matrix $A$ is positive semidefinite i.e. $x^TAx \geq 0$ if and only if all its eigenvalues are nonnegative.

Now suppose I have a matrix (not necessarily symmetric) $A$, whereby all its eigenvalues are nonnegative (obviously real), is it positive semidefinite?

My hunch is yes. Because such matrix $A$ would satisfy $\lambda_{\min}(A)\|x\|^2 \leq x^TAx$. Therefore it has to be positive semidefinite.

But I have searched up and down through every linear algebra book that I have came across, virtually all of them states definition with respect to symmetric positive semidefinite matrix only.

Answer 1

No, consider the following matrix:

$$A=\pmatrix{1&-10\\0&2}$$

It has positive eigenvalues, but is certainly not positive semidefinite.