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$A,B\in M_n$, $A$ is non-singular and $B$ is singular. $|||\cdot|||$ is any matrix norm on $M_n$, how to show that $|||A-B||| \geq \frac{1}{|||A^{-1}|||}$?

The hint is let $B=A[I-A^{-1}(A-B)]$, but I don't know how to use it.

Appreciate any help!

update: is $\geq$,not $\leq$.Sorry!

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I think you want the inequality $|||A - B ||| \ge \frac{1}{|||A^{-1}|||}$. The case $B = 0$ provides a counterexample. –  Hans Engler Nov 17 '12 at 14:12
    
@richard: Is it now? Consider $(A-B)v$ for $v\in \ker B$. –  tomasz Nov 17 '12 at 14:35
    
@tomasz: Thank you. Now I see. –  23rd Nov 17 '12 at 14:43
    
@HansEngler: After reading tomasz's comment, I finally realized that you are correct. I am sorry for bothering you. –  23rd Nov 17 '12 at 14:45
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3 Answers

up vote 2 down vote accepted

Hint for operator norm: notice that $\frac{1}{|||A^{-1}|||}=\inf_{v\neq 0} \lVert Av\rVert/\lVert v\rVert$, and then consider $(A-B)v$ for $v\in \ker B$.

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How can you argue that $\frac{1}{|||A^{-1}|||}=\inf_{v\neq 0} \lVert Av\rVert/\lVert v\rVert$? Similarly,if $f$ is a operator,shall we have this $\frac{1}{|||f^{-1}|||}=\inf_{v\neq 0} \lVert fv\rVert/\lVert v\rVert$ –  user39843 Nov 18 '12 at 16:35
    
@user39843: Straight from the definition. Yes, it is true for arbitrary (bounded, invertible) operators. –  tomasz Nov 18 '12 at 19:26
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This is wrong. In what follows, we assume that we are working with the operator norm. If $A$ is non-singular, it is easy to check that $P$ is non-singular for all $P\in B\left(A;\frac{1}{\Vert A^{-1}\Vert}\right)$.

The proof runs as follows. Let $P\in B\left(A,\frac{1}{\|A^{-1}\|}\right)$. We shall prove that $P\in GL_n(\mathbb{R})$. For each $x\in\mathbb{R}^n$, \begin{eqnarray*} \|x\|&=&\|A^{-1}(Ax)\|\leqslant \|A^{-1}\|\|Ax\|\\& \leqslant & \|A^{-1}\|\left(\|(A-P)x\|+\|Px\|\right)\leqslant \|A^{-1}\|\left(\|A-P\|\|x\|+\|Px\|\right). \end{eqnarray*} Thus, $$ \|x\|\leqslant \frac{\|A^{-1}\|}{1-\|A-P\|\|A^{-1}\|}\|Px\|,\text{ for all }x\in\mathbb{R}^n. $$ This implies that $P\in GL_n(\mathbb{R})$ because if $ Px=0$, we have $x=0$. Thus we have proved that $ B\left(A,\frac{1}{\|A^{-1}\|}\right)\subset GL_n(\mathbb{R})$.

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Let's sharpen the hint to $A^{-1}B = I - A^{-1}(A-B)$. First you should check that this identity is correct.

Now pick any $v$ such that $Bv = 0$ and $\|v \| = 1$. By assumption, such a $v$ exists. Apply both sides of the identity, play around with it, take norms, see if you can get something that resembles the statement that you want to prove.

Remember the definition of a matrix norm: $|||C||| = \sup_{\|x\| = 1} \|Cx\|$. There is also a formula that relates $|||CD|||$ to $|||C|||$ and $|||D|||$. Check your notes and try to use it.

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you say "By assumption,such a $v$ exists" which puzzles me. By what assumption? how can we argue that this $v$ exists? Thank you! –  user39843 Nov 17 '12 at 15:43
    
There is an assumption about $B$. –  Hans Engler Nov 17 '12 at 18:12
    
$|||A^{-1}B|||=|||I-A^{-1}(A-B)|||=max_{||v||=1}||[I-A^{-1}(A-B)]v||$ oh I fail to get it. I know that if I can argue that $||I||=1=||A^{-1}(A-B)\leq ||A^{-1}|| ||A-B||$ then I can solve the problem. But I cannot do that. –  user39843 Nov 19 '12 at 15:19
    
Since $B$ is singular, there is a $v \ne 0$ such that $Bv = 0$. Then $0 = A^{-1}Bv = v - A^{-1}(A-B)v$ and therefore $v = A^{-1}(A-B)v$. Now take norms, then $$\|v\| = \|A^{-1}(A-B)v\| \le |||A^{-1}||| \|(A-B)v\| \le |||A^{-1}||| |||A-B||| \|v\|$$. Divide by $\|v\| \ne 0$, then $|||A^{-1}||| |||A-B||| \ge 1$ and the proof is complete. –  Hans Engler Nov 21 '12 at 19:15
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