TITLE: Prove that every positive semidefinite matrix has nonnegative eigenvalues
QUESTION [26 upvotes]: There is a theorem which states that 

every positive semidefinite matrix only has eigenvalues $\ge0$

How can I prove this theorem?

REPLY [27 votes]: Recall the definition of an eigenvalue $\lambda$ (and an eigenvector $\vec{v}$):
$$A\vec{v}=\lambda\vec{v}$$
For a matrix to be positive semi-definite, $\vec{x}^TA\vec{x}\ge0$ for all $\vec{x}$. But if $\vec{v}$ is an eigenvector of $A$, then
$$\vec{v}^T A \vec{v} = \vec{v}^T (\lambda \vec{v}) = \vec{v}^T \vec{v} \lambda$$
Since $\vec{v}^T \vec{v}$ is necessarily a positive number, in order for $\vec{v}^TA\vec{v}$ to be greater than or equal to $0$, $\lambda$ must be greater than or equal to $0$.