1

Say I have a matrix of the form

$$ A = \begin{bmatrix} A_1 \\ A_2 \end{bmatrix} $$

where $A_1$ is a $n \times n$ matrix of full rank, and $A_2$ is a $(m-n) \times n$ matrix (arbitrary).

I would like to show that $||A^+||_2\leq||A_1^{-1}||_2$

where $A^{+}$ is the pseudo inverse, defined $(A^*A)^{-1}A^*$. I have said that, $$A^+ = (A_1^{*}A_1 + A_2^{*}A_2 )^{-1}[A_1^* A_2^*]$$

lets look at the first part of this matrix $$(A_1^{*}A_1 + A_2^{*}A_2 )^{-1}A_1^* = [A_1 + A_1^{-*}A_2^*A_2]^{-1}$$ this is as far as I can go with regards to manipulating this matrix as $A_2$ is not square it does not have inverse.

I am a bit unsure how to progress...

1 Answers1

0

Since $\|A_1^{-1}\|_2=\|(A_1^\ast)^{-1}\|_2$, it suffices to prove that for every nonzero vector $x\in\mathbb C^m$, there exists some nonzero vector $u\in\mathbb C^n$ such that $$ \|A^+x\|_2/\|x\|_2\le \|(A_1^\ast)^{-1}u\|_2/\|u\|_2.\tag{1} $$ $(1)$ is evident when $A^+x=0$. So, suppose $A^+x\ne0$. Pick $u=A^+x$. Then it suffices to show that $$ \|u\|_2^2\le \|x\|_2\|(A_1^\ast)^{-1}u\|_2.\tag{2} $$ Since $AA^+$ is an orthogonal projection, $\|x\|_2\ge\|AA^+x\|_2=\|Au\|_2\ge\|A_1u\|_2$. Hence it suffices to prove that $$ \|u\|_2^2\le \|A_1u\|_2\|(A_1^\ast)^{-1}u\|_2,\tag{3} $$ but this is just Cauchy-Schwarz inequality.

user1551
  • 139,064
  • what do you mean by $AA^+$ is an orthogonal projection? –  Aug 15 '17 at 19:23
  • @HMPARTICLE Take it literally. $P=AA^+$ is a projection (as $P^2=P$) and is actually an orthogonal projection (as $P$ is also Hermitian). – user1551 Aug 15 '17 at 19:30