4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

(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 $(A^+)^2=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.