Power of Moore-Penrose Inverse

Question

Let $A^+$ be the Moore-Penrose inverse of $A \in {\mathbb R}^{n\times n}$. Does the following hold?

$$(A^m)^+ = (A^+)^m, \quad m \in {\mathbb Z}_+.$$

Thanks!

Answer 1

No. E.g. when $A=A^2=\pmatrix{0&1\\ 0&1}$, we have \begin{aligned} A^+&=\frac12A^T,\\ \left(A^+\right)^2&=\frac14(A^T)^2=\frac14A^T,\\ \left(A^2\right)^+&=A^+=\frac12A^T.\\ \end{aligned}