If $A \in \mathcal{B}(H), B \in \mathcal{B}(K)$ are contractions, is it then true, that $\begin{bmatrix}A&0\\0& B \end{bmatrix}$ is a contraction too?

Question

$A \in \mathcal{B}(H)$ is a contraction, if:

$$\| A x \| \le \| x \|$$

for every $x \in H$. In other words: $\| A \| \le 1$.

The question here is: How can I show, that a matrix of operators (which itself is an operator) is a contraction?

The problem I'm having is what norm to choose from here. Should I use the Frobenius norm? Or another matrix-norm?

Answer 1

I assume that $\|(x,y)\|^2=\|x\|^2+\|y\|^2.$ Then $$\|(Ax,By)\|^2=\|Ax\|^2+\|By\|^2\\ \le \|x\|^2+\|y\|^2=\|(x,y)\|^2$$ Hence the operator is indeed a contraction. A similar argument works if we equip $H\times K$ with the norm of the form $\|(x,y)\|=\|x\|+\|y\|$