If a matrix $A$, not necessarily symmetric, has real, nonnegative eigenvalues, is it positive semidefinite? 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. 
 A: No, consider the following matrix:
$$A=\pmatrix{1&-10\\0&2}$$
It has positive eigenvalues, but is certainly not positive semidefinite. 
A: The answer to your question is NO. (And it is worse in the complex case: if a matrix $A$ is not Hermitian, then it is impossible that $x^TAx\geq 0$ for all $x$.)
In the setting of complex vector spaces, if $x^TAx\geq 0$ for all complex vectors $x$ (your definition of semi-definite), which in particular implies that $x^TAx\in{\bf R}$, the following theorem in Axler's Linear Algebra Done Right shows that $A$ must be Hermitian (if $A$ is real, then $A$ must be symmetric):

In the setting of real vector space, one has the following simple counterexample. 
For any $(x,y)\in{\bf R}^2$,
$$
(x,y)\begin{pmatrix}
0&1\\
0&0
\end{pmatrix}
\begin{pmatrix}
x\\y
\end{pmatrix}=xy. 
$$
A: 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.
