Singular Values/l2-norm of Pseudo-inverse 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 $y$.
