18
$\begingroup$

Show for any induced matrix norm and nonsingular matrix A that $$ \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1} $$ where $$ \left\|A^{-1}\right\| = \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}\\ \left\|A\right\| = \max_{\left\|x\right\|=1}\{\left\|Ax\right\|\}. $$

I am not sure how to show that: \begin{equation} \begin{split} \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1}\\ \text{or}\\ \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\} ≥ (\max_{\left\|x\right\|=1}\{\left\|Ax\right\|\})^{-1} \end{split} \end{equation}

$\endgroup$
4
  • $\begingroup$ i am still learning how to write equations in here, is there a faster way? Ill put what Ive tried in the actual question portion, it doesn't seem to be working here $\endgroup$
    – Megan
    Sep 24, 2014 at 1:41
  • 1
    $\begingroup$ Please don't vote this question down, I am still learning how to use the special text. Thanks. $\endgroup$
    – Megan
    Sep 24, 2014 at 1:53
  • $\begingroup$ @ Megan: welcome to MSE! $\endgroup$ Sep 24, 2014 at 2:11
  • $\begingroup$ Thanks! I used to use stack exchange all the time for coding, so excited to discover the math stack exchange. :) $\endgroup$
    – Megan
    Sep 24, 2014 at 2:18

3 Answers 3

33
$\begingroup$

Use $\lVert AB\rVert \leq \lVert A\rVert \lVert B\rVert$, as the induced norm is in particular submultiplicative. So that $\lVert I_n\rVert \leq \lVert A\rVert \lVert A^{-1}\rVert$.

$\endgroup$
5
$\begingroup$

Suppose $|y|=1$ is such that $|A|=|Ay|$. Then $x=Ay/|A|$ also has norm $1$ so it follows that $$ |A^{-1}|\geq |A^{-1}x|=|A^{-1}Ay/|A||=\frac{|y|}{|A|}=\frac{1}{|A|}=|A|^{-1}. $$

$\endgroup$
-1
$\begingroup$

Proof: $$ \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1} $$ where $$ \left\|A^{-1}\right\| = \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}\\ \left\|A\right\| = \max_{\left\|x\right\|=1}\{\left\|Ax\right\|\}\\ \left\|A\right\|^{-1} = (\max_{\left\|x\right\|=1}\{\left\|Ax\right\|\})^{-1} = (\min_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}) $$ and $$ \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\} ≥(\min_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}) $$ therefore: $$ \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1} $$

$\endgroup$
1
  • 3
    $\begingroup$ Are you supposing here that $\min_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\} = \|A\|^{-1}$? That's a very interesting supposition, and you should probably prove that as a part of your proof $\endgroup$ Sep 24, 2014 at 3:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .