I am trying to prove, given a matrix $A=\lbrack\frac{A_1}{A_2}\rbrack\in C^{m\times n}$, with $A_1\in C^{n\times n}$ non-singular, that:
$||A^+||_2\leq||A_1^{-1}||_2$
($||\cdot||_2$ is the induced $\ell_2$ norm, $(\cdot)^+$ is the Moore-Penrose pseudoinverse.)
Supposed to be simple but I'm having trouble relating $A$'s singular values to $A_1$'s. Any ideas?
When $A^+x$ is nonzero, so is $AA^+x$. Therefore \begin{aligned} \|A^+\|_2 &= \max\limits_{x\in\mathbb C^m\setminus0}\frac{\|A^+x\|_2}{\|x\|_2}\\ &\le\max\limits_{x\in\mathbb C^m\setminus0,\ A^+x\ne0}\frac{\|A^+x\|_2}{\|AA^+x+(I-AA^+)x\|_2}\\ &\le \max\limits_{x\in\mathbb C^m\setminus0,\ A^+x\ne0}\frac{\|A^+x\|_2}{\|AA^+x\|_2}\\ &\le \max\limits_{y\in\mathbb C^n\setminus0}\frac{\|y\|_2}{\|Ay\|_2}, \end{aligned} If we replace $A$ by $A_1$, all inequalities above become equalities. That is, $$ \|A_1^{-1}\|_2 = \max\limits_{x\in\mathbb C^n\setminus0}\frac{\|A_1^{-1}x\|_2}{\|x\|_2} = \max\limits_{x\in\mathbb C^n\setminus0}\frac{\|A_1^{-1}x\|_2}{\|A_1A_1^{-1}x\|_2} = \max\limits_{y\in\mathbb C^n\setminus0}\frac{\|y\|_2}{\|A_1y\|_2}. $$ The result now follows because $\|A_1y\|_2\le\|Ay\|_2$ for every vector $y$.
