Prove that if $T$ a normal linear transformation and invertible, then $T^{-1}$ is normal.

Question

The question is: Prove that if $T$ a normal linear transformation and invertible, then $T^{-1}$ is normal. Then I have to find the spectral decomposition of $T^{-1}$. At first I tried to prove it by definition: $T$ is normal so $TT^*=T^*T$ and I marked $S=T^{-1}$. So $ST=TS=I$.

Now I tried to prove that $SS^*=S^*S$ but didn't succeed by now. Many thanks to the helpers.

Answer 1

Set $S=T^{-1}$; then $S^*=(T^{-1})^*=(T^*)^{-1}$.

Therefore

$$ SS^*=T^{-1}(T^*)^{-1}=(T^*T)^{-1} $$

Can you go on from here?

For the spectral decomposition, observe that if $\lambda$ is an eigenvalue of $T$, then $\lambda\ne0$, so, for an eigenvector $v$ relative to $\lambda$ we have \begin{gather} Tv=\lambda v\\ STv=S(\lambda v)\\ v=\lambda Sv\\ Sv=\lambda^{-1}v \end{gather} and therefore $E_T(\lambda)=E_S(\lambda^{-1})$ (denoting with $E_A(\lambda)$ the eigenspace of the matrix $A$ relative to the eigenvalue $\lambda$). So it should be easy to write down the spectral decomposition of $S$.