# Prove that every positive semidefinite matrix has nonnegative eigenvalues

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 $$\vec{v}$$):

$$A\vec{v}=\lambda\vec{v}$$

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 $$0$$.

• This does not quite work. You need to argue first that the eigenvalues are real. Once you have that, your argument is fine. – 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 $x\in\mathbb{R}^n$.

See

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