There is a theorem which states that

every positive semidefinite matrix only has eigenvalues $\ge0$

How can I prove this theorem?


Recall the definition of an eigenvalue $\lambda$ (and an eigenvector $\vec{v}$):


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


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

Hint: Start with the definition. The $n\times n$ symmetric matrix $A$ is positive semidefinite if $x^TAx\geq 0$ for all $x\in\mathbb{R}^n$.




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


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