A linear operator is non-negative iff all its eigenvalues are non-negative

Question

I am asked to present a proof/ counterexample:

A linear operator $T$ on a finite-dimensional complex inner-product space is non-negative iff each of its eigenvalues are non-negative.

I think this claim is true. My thoughts:

Let $\lambda$ be an eigenvalue of $T$. Then for some nonzero $x$,
$$
\lambda \|x\|^2 = \langle \lambda x,x \rangle = \langle T x,x\rangle \ge 0.$$

Am I correct or did I miss something?

Thanks in advance

Answer 1

If $T$ is non-negative definite then all the eigen values of $T$ are non-negative, as you have shown. Converse is false. Consider the $2 \times 2$ matrix $a_{12}=1$ and $a_{ij}=0$ for all other $i,j$. Then the only eigen value is $0$ but this is not non-negative definite.