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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?

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Since $A$ has full column rank, we have $$ \|A^+\|_2 = \max\limits_{x\in\mathbb C^m\setminus0}\frac{\|A^+x\|_2}{\|x\|_2} = \max\limits_{y\in\mathbb C^n\setminus0}\frac{\|A^+Ay\|_2}{\|Ay\|_2} \le \max\limits_{y\in\mathbb C^n\setminus0}\frac{\|y\|_2}{\|Ay\|_2}. $$ Similarly, $$ \|A_1^{-1}\|_2 = \max\limits_{x\in\mathbb C^m\setminus0}\frac{\|A_1^{-1}x\|_2}{\|x\|_2} = \max\limits_{y\in\mathbb C^n\setminus0}\frac{\|A_1^{-1}A_1y\|_2}{\|A_1y\|_2} = \max\limits_{y\in\mathbb C^n\setminus0}\frac{\|y\|_2}{\|A_1y\|_2}. $$ The result now follows because $\|A_1y\|_2\le\|Ay\|_2$.

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  • $\begingroup$ @DushyantSahoo See my new edit. $\endgroup$ – user1551 Oct 29 '18 at 7:32

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