Does $(\mathbf{A}^{-1})^{H} = (\mathbf{A}^{H})^{-1}$ holds for complex-valued matrix?

Question

Does $(\mathbf{A}^{-1})^{H} = (\mathbf{A}^{H})^{-1}$ holds for complex-valued matrix?

Based on my quick proof it seems like this will hold but I am not sure if there is anything about complex-valued matrices that I am missing here. I am personally not too familiar with complex-valued matrices. \begin{align} (\mathbf{A}^{-1})^{H}\mathbf{A}^H = (\mathbf{AA}^{-1})^H = I^H = I\\ \mathbf{A}^H\mathbf{A}^{-1})^{H} = (\mathbf{A^{-1}}\mathbf{A})^H = I^H = I\\ \end{align}

Answer 1

Yes, it holds for complex matrices. And your proof is correct, although too long. If you have two square matrices of the same size $A$ and $B$, then, in order to prove that $B=A^{-1}$, it is enough to prove that one of these assertions hold:

• $AB=\operatorname{Id}$;
• $BA=\operatorname{Id}$.

In fact, if one of them holds, then the other one will hold too.