Showing the adjoint of $A = A^{-3}$ iff $A^* = A$ and $A^2 = I$

I'm trying to show that $$A^* = A^{-3}$$ iff $$A^* = A$$ and $$A^2 = I$$ given $$A$$ is invertible and $$A\in M_n(\mathbb{C})$$ but I'm having trouble with manipulating $$A$$ and just don't know what's legal or not. $$A^*$$ here is the adjoint of $$A$$. Any help would be appreciated!

I believe it's possible to do just by algebraic manipulation on $$A$$, but super lost as to how to achieve it.

Just to be super clear. $$A^*$$, the adjoint of $$A$$ is the transpose of the conjugate of $$A$$.

• Do you mean the Adjugate with $A^*$? Usually, $A=A^*$ says that $A$ is normal, and $A^*$ is the conjugate transpose. So then just $A=A^{-3}$? Please clarify the assumptions. Nov 12 '21 at 16:04
• Yes $A^*$ here refers to the transpose of the conjugate of $A$. Sorry if the notation wasn't clear Nov 12 '21 at 16:11
• $\Leftarrow$ part is so trivial, because if $A^*=A\ \wedge\ A^2=I$ , then $A^{-3}=IIA^{-3}$. So... the question is prooving $\Rightarrow$ part. Nov 12 '21 at 16:15
• Yes agreed, the backwards direction I have shown. I have a solution for the forward direction, but the solution uses unclear/perhaps illegal matrix algebra, so I'm not confident in it as well. Nov 12 '21 at 16:17

If $$A^* = A^{-3}$$ then $$A^*A = AA^*$$ ($$A$$ is normal), so by the Spectral Theorem, $$A$$ is diagonalizable. Do a change of basis so that $$A$$ is diagonal. And now you can check that the only complex numbers that satisfy $$\bar\lambda = \lambda^{-3}$$ are $$\lambda = \pm 1$$.
If $$A$$ is diagonal and the diagonal entries are $$\pm 1$$ then $$A^* = A$$ and $$A^2 = I$$.
Since $$A^\ast=A^{-3}$$, we see that $$A^\ast$$ commutes with $$A$$. Hence $$A^2=(A^\ast A)^{-1}$$ is self-adjoint and positive definite. It follows that $$A^2=(A^2)^\ast=(A^\ast)^2=(A^{-3})^2$$, meaning that $$A^8=I$$ or $$(A^2-I)(A^2+I)(A^4+I)=0.\tag{1}$$ Since $$A^2$$ is positive definite, so are $$A^2+I$$ and $$A^4+I$$. Therefore $$(1)$$ implies that $$A^2-I=0$$. But then we also have $$AA^\ast-I=0$$ because $$A^\ast=A^{-3}=(A^2)^{-1}A^{-1}=A^{-1}$$. Hence $$A=A^\ast$$.