I am trying to write up solutions for my linear algebra students and I'm currently stuck on the following proof:

Let $A$ be a square matrix such that $A^2 = O$. Prove that $(A^\dagger)^2 = O$.

First off, the students do not know the Penrose conditions of the pseudoinverse. In particular, they have that for an $m\times n$ matrix $A$, the pseudoinverse $A^\dagger$ is given in terms of the singular value decomposition of $A$. That is, if $$A = U\Sigma V^*$$ for unitary matrices $U$ and $V$, then $$A^\dagger = V\Sigma^\dagger U^*$$ where $\Sigma^\dagger$ is the $n\times m$ matrix with $1/\sigma_i$ along the diagonal (for $\sigma_i$ the $i^{\text{th}}$ non-zero singular value of $A$) and zero elsewhere.

My gut tells me that the polar decomposition of a square matrix

$$A = WP$$ for some unique matrices $W$ (unitary) and $P$ (positive semi-definite)

will be important here. They have also proved in a previous problem that

$$(UA)^\dagger = A^\dagger U^*$$ $$(AU)^\dagger = U^* A^\dagger$$ for any $A$ and any $U$ unitary

which I think should come into play here. Any help getting to the end of of this proof would be much appreciated. I feel like I'm almost there, I just can't see it right now.


(I write $A^+$ for $A^\dagger$.) I think it's easier to prove as follows. Suppose $A=U(S\oplus0)V^\ast$ is a SVD, where $S$ is invertible. Partition $V^\ast U$ as $\pmatrix{X&Y\\ Z&W}$ where $X$ has the same size as $S$. Then \begin{align*} A^2&=U\pmatrix{S\\ &0}\pmatrix{X&Y\\ Z&W}\pmatrix{S\\ &0}V^\ast =U\pmatrix{SXS&0\\ 0&0}V^\ast,\\ (A^+)^2&=V\pmatrix{S^{-1}\\ &0}\pmatrix{X^\ast&Y^\ast\\ Z^\ast&W^\ast}\pmatrix{S^{-1}\\ &0}U^\ast =V\pmatrix{S^{-1}X^\ast S^{-1}&0\\ 0&0}U^\ast. \end{align*} Hence $A^2=0$ implies that $X=0$ and in turn $(A^+)^2=0$.


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