Relation between operator norm of a matrix and norm of inverse

Question

Given a matrix $A$ over $\mathbb{R}$, define the operator norm as $\|A\|: = \sup\{\|A\mathbf{x}\| : \|\mathbf{x}\|=1\}$.

If $A$ is invertible, I realize that in general we have $\|A^{-1}\|\|A\|\geq 1$.

My question:

What is a concrete example where this inequality is strict? I can't think of one.

@angryavian That question has no information on producing a strict inequality. — CuriousKid7, May 03 '17 at 03:54
Try $A= \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$ or $A= \begin{bmatrix} 1 & 0 \ 0 & {1 \over 2} \end{bmatrix}$ . — copper.hat, May 03 '17 at 04:11

score 1 · Answer 1 · answered May 03 '17 at 04:14

1

$(\mathbb{R}^2, \|\cdot\|_{sup})\to(\mathbb{R}^2, \|\cdot\|_1)$ i.e. $\|(x,y)\|_1=|x|+|y|$

$A= \left[ {\begin{array}{cc} 0 & 1\\ 1 & 0 \\ \end{array} } \right]=A^{-1}$

We have $\|(1,1)^{t}\|_{sup}=1$. Thus $\|A(1,1)^{t}\|_1=\|(1,1)^{t}\|_1=2$ then $\|A\|=\|A^{-1}\|>1$. Finally $\|A\|\cdot\|A^{-1}\|>1$.

answered May 03 '17 at 04:14

freshman

1 Answers1