0

Let $A$ be an $m\times n$ matrix with coefficients in $\mathbb R$. Supposedly this fact is true, however when I compute it:

\begin{align} (A^+ A)^\ast &= A^\ast (A^+)^\ast \newline&= A^\ast ((A^\ast A)^{-1}A^\ast)^\ast \newline &= A^\ast A(A^\ast A)^{-1} \end{align}

Which is clearly not equal to $A^+ A$.

jem do
  • 305
  • 1
    This is true even if $A$ is not of full rank. – Mittens Apr 12 '23 at 06:03
  • 1
    clearly not equal? Well, I'd said it's clearly the identity matrix, so... (of course when the matrix inverse makes sense, which you are assuming here) – Jean-Claude Arbaut Apr 12 '23 at 07:12
  • It appears that you define $A^+$ as $(A^\ast A)^{-1}A^\ast$ (note that the inverse makes sense only when $A$ has full column rank). In this case $A^\ast A=(A^\ast A)^{-1}A^\ast A=I$. Therefore $(A^\ast A)^\ast=I^\ast=I=A^\ast A$. – user1551 Apr 12 '23 at 09:38

2 Answers2

2

What's your definition of the pseudoinverse? Classically in more general case of any complex matrix, $A^+$ by definition is a unique matrix $A^+$ satisfying all of the four conditions:

  • $AA^+A = A$
  • $A^+AA^+ = A^+$
  • $(A^+A)^* = A^+A$
  • $(AA^+)^* = AA^+$

So there is nothing to show. In case where $A$ has linearly independent columns, we get a more compact formula that you used $A^+ = (A^*A)^{-1}A^*$

In this case the fact is trivial, as $$(A^*A)^{-1}A^*A = I$$

Linear independence of columns is a necessary condition for the invertibility of $A^*A$. Similar formula can be derived in case of linear independent rows, as then $AA^*$ is non-singular and $$A^+ = A^*(AA^*)^{-1}$$

In this case $$A^+A = A^*(AA^*)^{-1}A = (A^{*}(A^{*}(AA^{*})^{-1})^*)^* = (A^* (AA^* )^{-1} A)^* = (A^+A)^*$$

Alexander Golys
  • 829
0

The singular value decomposition makes the problem related to the pseudoinverse simpler in many cases. Let $A = U\Sigma V^*\in\mathbb{C}^{m\times n}$ be the singular value decomposition of a matrix $A$ where first $r={\rm rank}A$ diagonal entries of $\Sigma$ are nonzero singular values $\sigma_1,\dots,\sigma_r$. Then, its pseudoinverse can be written as $A^+ = V\Sigma^+U^*$ where first $r$ diagonal entries of $\Sigma^+$ are $1/\sigma_1,\dots,1/\sigma_r$. Thus, we can write $A^+A = V\Sigma^+\Sigma V^*$ and easily check $\Sigma^+\Sigma = (\Sigma^+\Sigma)^*\in\mathbb{C}^{n\times n}$. Therefore, $(A^+A)^* = (V\Sigma^+\Sigma V^*)^* = V\Sigma^+\Sigma V^* = A^+A$.

Indeed, this is a part of the proof to show that the pseudoinverse can be written by using the singular value decomposition. We can easily check the other three properties (see Alexander's answer) similarly.

flyingwith
  • 902