Why $\left\| \begin{pmatrix} A &0\\ 0 &0 \end{pmatrix}\right\|=\|A\|$?

Question

Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all operators on $E$.

Let $A\in\mathcal{L}(E)$. I want to prove that $$\left\| \begin{pmatrix} A &0\\ 0 &0 \end{pmatrix}\right\|=\|A\|.$$ Here $\begin{pmatrix} A &0\\ 0 &0 \end{pmatrix}\in \mathcal{L}(E\oplus E)$.

It is well-known that $$\left\| \begin{pmatrix} A &0\\ 0 &0 \end{pmatrix}\right\|\leq \left\| \begin{pmatrix} \|A\| &0\\ 0 &0 \end{pmatrix}\right\|=\|A\|.$$ I'm facing difficulties to prove the converse inequality.

Answer 1

Since $\begin{bmatrix} A & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\y\end{bmatrix} = \begin{bmatrix} Ax \\ 0\end{bmatrix}$ and $\|\begin{bmatrix} Ax \\ 0\end{bmatrix} \|= \|Ax\|$, the result follows immediately from the definition of the operator norm.

Answer 2

The equality can be directly proved \begin{align*} \left\| \begin{pmatrix} A &0\\ 0 &0 \end{pmatrix}\right\| &= \sup_{\substack{a\in E^2\\\|a\|^2=1}} \left\| \begin{pmatrix} A &0\\ 0 &0 \end{pmatrix}\cdot a\right\|\\ &= \sup_{\substack{x\in E,y\in E\\\|x\|^2+\|y\|^2=1}} \left\| \begin{pmatrix} A &0\\ 0 &0 \end{pmatrix}\cdot\begin{pmatrix} x\\y \end{pmatrix}\right\|\\ &= \sup_{\substack{x\in E\\\|x\|^2=1}} \left\| \begin{pmatrix} A &0\\ 0 &0 \end{pmatrix}\cdot\begin{pmatrix} x\\0 \end{pmatrix}\right\|\\ &= \sup_{\substack{x\in E\\\|x\|^2=1}} \left\| A\cdot x\right\|\\ &=\|A\| \end{align*}