# 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?

## 3 Answers

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 $0$.

• 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.