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