TITLE: If a matrix $A$, not necessarily symmetric, has real, nonnegative eigenvalues, is it positive semidefinite?
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: Given that, for a real matrix $A$ we have
$$
x^T A x=x^T A^{(S)} x
$$
where $A^{(S)}$ is the symmetric part of $A$, then the character of $A$ with respect to positive definiteness, or semi-definiteness, is related only to the eigenvalues of its symmetric part. 
As the examples in other answers show, a matrix could have positive eigenvalues, but its symmetric part could have a negative eigenvalue, so eigenvalues of a matrix could not be related to positive (semi)definiteness.