Showing a linear operator in block matrix form has a specific operator norm.

Question

I'm trying to show the following:

Let $T$ be a square matrix, and suppose both subspaces $M$ and $M^{\perp}$ are $T$-invariant, so that $T$ takes the schematic form $\begin{pmatrix} A & 0\\ 0 & B \end{pmatrix}$. Show that $\| T\|=\max(\|A\|,\| B\|)$.

I get to a certain step using the hint some of the hints in the book, but I'm stuck at where to go next. Here's my work so far:

Let $x=a+b$ where $a\in M$ and $b\in M^{\perp}$. Then, $Tx=Ta+Tb$, where $Ta=Aa\in M$ and $Tb=Bb\in M^{\perp}$. Since $\langle a,b\rangle=0$ and $\langle Aa,Bb\rangle=0$, by Pythagoras we have $$\|T\|^2=\sup \frac{\|Tx\|^2}{\|x\|^2}=\sup \frac{\|Ta\|^2+\|Tb\|^2}{\|a\|^2+\|b\|^2}. $$ Further, $$ \frac{\|Ta\|^2+\|Tb\|^2}{\|a\|^2+\|b\|^2}=\frac{\|Aa\|^2+\|Bb\|^2}{\|a\|^2+\|b\|^2}\leq \dfrac{\|A\|^2\|a\|^2+\|B\|^2\|b\|^2}{\|a\|^2+\|b\|^2}$$ and setting $t=\dfrac{\|a\|^2}{\|a\|^2+\|b\|^2}$ gives us $$\|T\|^2\leq t\|A\|^2+(1-t)\|B\|^2. $$ If $\|A\|\leq \|B\|$, then setting $t=0$ (so $\|a\|^2=0$) gives us $\|T\|^2\leq \|B\|$, and if $\|B\|\leq \|A\|$, then setting $t=1$ (so $\|b\|^2=0$) gives us $\|T\|^2\leq \|B\|$.

I'm not sure if this is the right approach, or how to get the equality from here.

Answer 1

To continue where you left off, \begin{align*} \|T\|^2&\leq t\|A\|^2+(1-t)\|B\|^2 \\ &\leq t\left(\max\{\|A\|^2,\|B\|^2\}\right)+(1-t)\left(\max\{\|A\|^2,\|B\|^2\}\right) \\ &=\max\{\|A\|^2,\|B\|^2\} \\ &=\left(\max\{\|A\|,\|B\|\}\right)^2, \end{align*} and thus $\|T\|\leq\max\{\|A\|,\|B\|\}$.
To prove the reverse inequality, first assume that $\|B\|\leq\|A\|$. Given $\varepsilon>0$ there is some $x\in M$ such that $\|x\|\leq1$ and $\|Ax\|\geq\|A\|-\varepsilon$. Thus, we have $$\max\{\|A\|,\|B\|\}-\varepsilon=\|A\|-\varepsilon\leq\|Ax\|=\|Tx\|\leq\|T\|.$$ Since $\varepsilon>0$ was arbitrary we have $\max\{\|A\|,\|B\|\}\leq\|T\|$, and thus $\|T\|=\max\{\|A\|,\|B\|\}$. The case that $\|A\|\leq\|B\|$ is handled in the same manner.