Proof explanation related to the operator matrices

functional-analysishilbert-spacesoperator-theory

Let $F$ be a complex Hilbert space. Let $A,B,C,D\in \mathcal{B}(F)$. Consider the operator matrix $T$ such that
\begin{equation*}
T=\begin{pmatrix}A & B \\
C & D
\end{pmatrix}\in \mathcal{B}(F\oplus F).
\end{equation*}
Consider
\begin{equation*}
\widetilde{T}=\begin{pmatrix}\|A\| & \|B\| \\
\|C\| & \|D\|
\end{pmatrix}.
\end{equation*}

I want to understand why
$$r(T)\leq r (\widetilde{T})?$$

Note that the inner product on $F\oplus F$ is defined as follows: If $x=\begin{pmatrix} x_1\\ x_2\end{pmatrix}\in F\oplus F$ with $x_1,x_2\in F$, and $x'=\begin{pmatrix}x_1'\\ x_2'\end{pmatrix}$ similarly, then
$$\langle x,x'\rangle_{F\oplus F}:= \langle x_1,x_1'\rangle_F +\langle x_2,x_2'\rangle.$$

I see the following theorem in a more general context but I'm facing difficulties to understand the proof

Theorem 1.1
Let $A = (A_{ij})_{n \times n}$ be an operator matrix and $\tilde{A} = (\| A_{ij} \|)_{n \times n}$ its block-norm matrix.
Then

  1. $\omega(A) \leq \omega(\tilde{A})$,
  2. $\| A \| \leq \| \tilde{A} \|$,
  3. $r_\sigma(A) \leq r_\sigma(\tilde{A})$.

(Original image here.)

Proof:
3. Notice, in general, that for operators $A = (A_{ij})_{n \times n}$ and $B = (B_{ij})_{n \times n}$ we have
$$
\| \widetilde{AB} \|
\leq \| \tilde{A} \tilde{B} \| .
$$
In fact, $\tilde{A} \tilde{B} – \widetilde{AB}$ is a nonnegative matrix since $\widetilde{AB} = (\| \sum_{k=1}^n A_{ik} B_{kj} \|)_{n \times n}$ and $\tilde{A} \tilde{B} = (\sum_{k=1}^n \| A_{ik} \| \| B_{kj} \|)_{n \times n}$.
Therefore, by the norm monotonicity of nonnegative matrices, we get $\| \widetilde{AB} \| \leq \| \tilde{A} \tilde{B} \|$.
Using induction, we have
$$
\| A^m \|
\leq \| \widetilde{A^m} \|
\leq \| \smash{\tilde{A}}\vphantom{A}^m \| ,
$$
for every positive integer $m$.
This leads to
$$
r_\sigma(A)
\leq r_\sigma(\tilde{A}).
$$

(Original images here and here.)

Best Answer

The steps are as follows:

  • Show that if $A,B$ are non-negative and $A\leq B$ (entrywise), then $\|A\|\leq \|B\|$. This is easy: since all entries of $A$ are $\geq0$, the max in $$\|A\|=\max\{ \|Ax\|:\ \|x\|=1\}$$ is achieved when all entries of $x$ are $\geq0$. Now, for any $x$ with $x_j\geq0$ for all $j$, $$ \|Ax\|^2=\sum_k\left(\sum_j A_{kj}x_j\right)^2\leq\sum_k\left(\sum_j B_{kj}x_j\right)^2=\|Bx\|^2 $$ (using essentially that all entries are positive). Thus $\|A\|\leq\|B\|$.

  • Show that $\tilde A\tilde B-\tilde{AB}$ is non-negative. This is simply done by looking, block by block, at $$ (\widetilde{AB})_{kj}=\left\|\sum_\ell A_{k\ell} B_{\ell j} \right\|\leq \sum_\ell \|A_{k\ell} \|\,\|B_{\ell j}\|=(\tilde A\tilde B)_{kj}. $$ Now, the previous step implies that $$\tag1 \|\widetilde {AB}\|\leq\|\tilde A\tilde B\|.$$

  • We have $$\tag2\|A\|\leq\|\tilde A\|$$ for any non-negative $A$. This follows from, when $\|x\|=1$,
    $$ \|Ax\|^2=\sum_k\left|\sum_jA_{kj}x_j\right|^2\leq\sum_k\left(\sum_j|A_{kj}|\,|x_j|\right)^2 \leq\sum_k\left(\sum_j(\tilde A)_{kj}\,|x_j|\right)^2\leq\|\tilde A\|. $$

  • Apply $(1)$ and $(2)$ inductively to get $$\tag3\|A^m\|\leq \|\widetilde{A^m}\|\leq\|(\tilde A)^m\|,\ \ \ m\in\mathbb N.$$

  • Now use $$ r(A)=\lim_m\|A^m\|^{1/m}\leq\lim_m\|(\tilde A)^m\|^{1/m}=r(\tilde A). $$

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