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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$\begingroup$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 $0$.
$\endgroup$ 1 $\begingroup$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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The first characterization (modified a bit for the semidefinite case) is what you want.
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