Hermitian Matrix eigenvalues

Question

I am trying to show the following sentence: Let H be hermitian with $\sigma(H) \subseteq \lbrace-r,r\rbrace$. Show that H ∘ H = $r^2$* I.

Since H is hermitian, we know that we can decompose the matrix into H=SDS* with S unitary and D diagonal matrix. H ∘ H = $(SDS^*) ∘ (SDS^*) = SD(S^* S)DS^* = SD^2S^*$. I know since the eigenvalues are $\subseteq \lbrace -r,r\rbrace, D^2 = r^2*I $ but that's not the fact that I needed to show. Any ideas where I am wrong?

($\sigma(H)$ is the set of all eigenvalues of H)

Answer 1

$S$ is unitary so $SS^\ast = I$. Thus $$H^2 = SD^2S^\ast = S(r^2 I)S^\ast = r^2 I SS^\ast = r^2I$$