On eigenvalues, hermitian matrices and SVD

Are my ideas on the following "true or false"-statements correct?

If $A$ is hermitian and $\lambda$ is an eigenvalue of $A$, then $|\lambda|$ is a singular value of $A$.

My answer would be yes, because we can write $D=P^{-1}AP$ where $D$ is the diagonal matrix (which contains $\lambda$) and $P$ is an invertible matrix. Next we could bring $P$ to the left side of the equation to get our SVD composition...

If $A$ is diagonalizable and all its eigenvalues are equal, then $A$ is diagonal.

I don't have a proof for this statement to be yes, but I can't find any contradictory examples as well...

1) Recall that the singular values of $A$ are precisely the square roots of the eigenvalues of $A^\ast A$; since $A$ is Hermitian, we have that $A^\ast A = A^2$. Given that you can diagonalise $A$ as $A = PDP^{-1}$ (where $P$, in fact, can be taken to be unitary), how can you immediately diagonalise $A^\ast A = A^2$?
2) If $A$ is diagonalisable, then $A = PDP^{-1}$ for $D$ diagonal and $P$ invertible. If all the eigenvalues are the same, i.e., $\lambda$, then $D = \lambda I$. What does this imply about $A = PDP^{-1}$?
• 1) $A^*A=A^2=PDP^{-1}PDP^{-1}=PD^2P^{-1}$, so yes, $|\lambda|$ is a singular value of $A$ because $\lambda^2$ is an element on the diagonal of $D$ and we have to take the root to get the singular value. – ABC Nov 7 '13 at 20:39
• 2) $A=PDP^{-1}=P \lambda I P^{-1}=\lambda PP^{-1}=\lambda I$, so yes $A$ is diagonal. – ABC Nov 7 '13 at 20:42