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There is a theorem which states that

every positive semidefinite matrix only has eigenvalues $\ge0$

How can I prove this theorem?

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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 $\vec{x}$ is an eigenvector of $A$, then

$$\vec{v}^T\vec{v}\lambda$$

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 $0$.

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    $\begingroup$ This does not quite work. You need to argue first that the eigenvalues are real. Once you have that, your argument is fine. $\endgroup$ – Andrés E. Caicedo Mar 7 '14 at 16:46
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Hint: Start with the definition. The $n\times n$ symmetric matrix $A$ is positive semidefinite if $x^TAx\geq 0$ for all $x\in\mathbb{R}^n$.

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See

http://en.wikipedia.org/wiki/Positive-definite_matrix#Characterizations.

The first characterization (modified a bit for the semidefinite case) is what you want.

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