3
$\begingroup$

If we have an invertible operator $B$ from $\mathbb{R^2}$ to $\mathbb{R^2}$, is it necessarily true that $\|B^{-1}\|= 1/\|B\|$?

I think the answer is no. We are on Rudin's Principle's of Mathematical Analysis's Chapter 9.

Using Theorem 9.7, which states if $A\in L(\mathbb{R^n},\mathbb{R^m})$ and $B\in L(\mathbb{R^m},\mathbb{R^k})$, then $$\|BA\| \leq \|B\| \|A\|.$$

Thus we have: $\|1\|=\|B^{-1}*B\|\leq \|B^{-1}\|\|B\|$. Thus $\|1\|/\|B\| \leq \|B^{-1}\|$. So we have $\|B^{-1}\| \geq \|1\|/\|B\|$.

Thus I get that $\|B^{-1}\|$ can be greater than or equal to $1/\|B\|$. However, I'm not sure if $B$ being an invertible operator specifically from $\mathbb{R^2}$ to $\mathbb{R^2}$ makes it just equal to (and not necessarily greater than. Thanks for the help.

$\endgroup$
0
3
$\begingroup$

$$\left\|\pmatrix{1&0\\0&2}\right\|=2,\quad\left\|\pmatrix{1&0\\0&1/2}\right\|=1$$

$\endgroup$
8
  • $\begingroup$ Thanks. This is what I thinking too. But can I consider $M_2(R)$ to $M_2(R)$ when we are considering $R^{2}$ to $R^{2}$? Or in other words, are these equivalent? $\endgroup$ – kemb Mar 9 '18 at 1:30
  • $\begingroup$ No, linear maps $\Bbb R^2\to\Bbb R^2$ themselves correspond to $2\times2$ matrices. $\endgroup$ – Berci Mar 9 '18 at 1:33
  • $\begingroup$ Oh I see, great thanks, I was slightly confused. $\endgroup$ – kemb Mar 9 '18 at 1:34
  • $\begingroup$ Last question, this may be a very simple question, but I never learned this: how do we calculate the norm of a matrix, i.e. why does the norm of the matrix on the right equal 1? Thanks! $\endgroup$ – kemb Mar 9 '18 at 1:36
  • 1
    $\begingroup$ @kemb: A simple inequality will show you that $\|Ax\| \le \|x\|$, and equality is attained for $x = (1,0)$. Similarly, you can conclude that for any diagonalizable matrix, the operator norm equals the absolute value of the largest eigenvalue: $\|A\| = \max_i |\lambda_i|$. $\endgroup$ – Nate Eldredge Mar 9 '18 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.