Let $A \in M_{n×n}\mathbb{(C)}$ be invertible. Then
$(A^∗)^{−1} = (A^{−1})^∗$
Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).
I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.
Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.
Best Answer
Just simply use the definition of $A$ in terms of $B$.
If you have $A=B^{-1}B^*$, then $$A^{-1}=B^{-*}B, \quad A^*=BB^{-*}.$$ So $$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$ and hence $A^{-1}$ and $A^*$ are similar.