Show that $0$ is the only eigenvalue of $Q$ given that $Q^*=5Q$

Question

$Q$ is a linear operator from $V \to V$ with $V$ being a finite dimensional complex inner-product-space.

Given: $Q^*=5Q$, $Q^*$ being the adjoint.

Show that $0$ is the only eigenvalue of $Q$.

I've been staring at this problem for quite a while now. I think the answer lies in the eigenvalues of $Q$ and $Q^*$ being the same, but I can't think of a way to prove $0$ is an eigenvalue let alone the only one.

Answer 1

Try doing this by direct computation. Suppose we have an eigenvalue $\lambda$ with (unit) eigenvector $v$. Then since we have a question about adjoints, it's natural that inner products will be used somewhere. Try considering $v$:

$$\langle Qv, v\rangle = \langle \lambda v, v \rangle = \lambda$$

On the other hand,

\begin{align*} \langle Qv, v\rangle = \langle v, Q^* v\rangle = \langle v, 5Q v \rangle = 5\overline{\lambda} \end{align*}

The result now follows.

Answer 2

If you have one nonzero eigenvalue, then its multiples by all powers of $5$ must also be eigenvalues.