# Is it necessarily true that $\|B^{-1}\|$= $1/\|B\|$?

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.

$$\left\|\pmatrix{1&0\\0&2}\right\|=2,\quad\left\|\pmatrix{1&0\\0&1/2}\right\|=1$$
• 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? – kemb Mar 9 '18 at 1:30
• No, linear maps $\Bbb R^2\to\Bbb R^2$ themselves correspond to $2\times2$ matrices. – Berci Mar 9 '18 at 1:33
• @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: $\|A\| = \max_i |\lambda_i|$. – Nate Eldredge Mar 9 '18 at 1:46