Proposition 40.5.4 | Jerry's Digital Garden

Proposition 40.5.4.label Let $a, b, c \in \complex$ with $a \ne b$, and

\[T = \begin{bmatrix}a &c \\ 0 &b\end{bmatrix}\]

then:

(1)
For each $p \in \complex[z]$,
\[p(T) = \begin{bmatrix}p(a) &c \frac{p(a) - p(b)}{a - b} \\ 0 & p(b)\end{bmatrix}\]
(2)
If $|a|, |b| \le 1$, then for each $f \in A(D)$,
\[f(T) = \begin{bmatrix}f(a) &c \frac{f(a) - f(b)}{a - b} \\ 0 & f(b)\end{bmatrix}\]
(3)
Suppose that $|a|, |b| < 1$ and let
\[\varphi: \complex \setminus \bracsn{b/|b|^2}\to \complex \quad \varphi(z) = \frac{z - b}{1 - \ol b z}\]

then $\norm{T}_{M_2(\complex)}\le 1$ if and only if
\[|\varphi(a)|^{2}(|a - b|^{2} + |c|^{2}) \le |a - b|^{2}\]
(4)
If $|a|, |b| \le 1$, then $\norm{T}_{M_2(\complex)}\le 1$ if and only if
\[|\varphi(a)|^{2}(|a - b|^{2} + |c|^{2}) \le |a - b|^{2}\]

Proof. (1): Suppose that (1) holds for $p, q \in \complex[z]$, then

\begin{align*}p(T)q(T)&= \begin{bmatrix}p(a) &c \frac{p(a) - p(b)}{a - b} \\ 0 & p(b)\end{bmatrix}\begin{bmatrix}q(a) &c \frac{q(a) - q(b)}{a - b} \\ 0 & q(b)\end{bmatrix} \\&= \begin{bmatrix}p(a)q(a)&c p(a)\frac{q(a) - q(b)}{a - b}+ cq(b)\frac{p(a) - p(b)}{a - b}\\ 0&p(b)q(b)\end{bmatrix}\\&= \begin{bmatrix}p(a)q(a) &c \frac{p(a)q(a) + p(b)q(b)}{a - b} \\ 0 & p(b)q(b)\end{bmatrix}\end{align*}

As (1) already holds for $1, z \in \complex[z]$, by induction, (1) holds for all $p \in \complex[z]$.

(3): For each $z \in \partial B_{\complex}(0, 1)$,

\[|\varphi(z)| = \abs{\frac{z - b}{1 - \ol{b}z}}= \abs{\frac{z - b}{\ol{z} - \ol{b}}}= 1\]

By the Maximum Modulus Theorem, $\varphi \in A(D)$ with $\norm{\varphi}_{A(D)}= 1$. By (2),

\[\varphi(T) = \begin{bmatrix}\varphi(a) & c\frac{\varphi(a)}{a - b} \\ 0 & 0\end{bmatrix} \quad \norm{\varphi(T)}_{M_2(\complex)}^{2} = |\varphi(a)|^{2} + \abs{c\frac{\varphi(a)}{a - b}}^{2}\]

so $\norm{\varphi(T)}_{M_2(\complex)}\le 1$ if and only if

\begin{align*}1&\ge |\varphi(a)|^{2} \paren{1 + \frac{|c|^{2}}{|a - b|^{2}}}\\ |a - b|^{2}&\ge |\varphi(a)|^{2} (|a - b|^{2} + |c|^{2})\end{align*}

(3, $\Rightarrow$): Suppose that $\norm{T}_{M_2(\complex)}\le 1$. By continuity of the holomorphic functional calculus and von Neumann’s inequality, $\norm{\varphi(T)}_{M_2(\complex)}\le 1$, and $|a - b|^{2} \ge |\varphi(a)|^{2} (|a - b|^{2} + |c|^{2})$.

(3, $\Leftarrow$): Suppose that $|a - b|^{2} \ge |\varphi(a)|^{2} (|a - b|^{2} + |c|^{2})$, or equivalently that $\norm{\varphi(T)}_{M_2(\complex)}\le 1$.

Let $\psi(z) = \frac{z + b}{1 + \ol b z}$, then $\psi \circ \varphi = \text{Id}$. For each $z \in \partial B_{\complex}(0, 1)$,

\[|\psi(z)| = \abs{\frac{z + b}{1 + \ol b z}}= \abs{\frac{z + b}{\ol z + \ol b}}= 1\]

By the Maximum Modulus Theorem, $\psi \in A(D)$ with $\norm{\psi}_{A(D)}= 1$. Since $\norm{\psi(T)}_{M_2(\complex)}\le 1$, the holomorphic functional calculus and von Neumann’s inequality imply that

\[\norm{T}_{M_2(\complex)}= \norm{\psi(\varphi(T))}_{M_2(\complex)}\le \norm{\psi}_{A(D)}= 1\]

(4): For each $t \in (0, 1)$, let $\varphi_{t}(z) = (z - tb)/(1 - t\ol b z)$.

If $\norm{T}_{M_2(\complex)}\le 1$, then $\norm{tT}_{M_2(\complex)}\le 1$ for all $t \in (0, 1)$, so

\begin{align*}|\varphi(a)|^{2}(|a - b|^{2} + |c|^{2})&= \lim_{t \upto 1}|\varphi_{t}(ta)|^{2}(|ta - tb|^{2} + |tc|^{2}) \\&\le \lim_{t \upto 1}|ta - tb|^{2} = |a - b|^{2}\end{align*}

On the other hand, if $|\varphi(a)|^{2}(|a - b|^{2} + |c|^{2}) \le |a - b|^{2}$, then

\begin{align*}|1 - t^{2}\ol b a|^{2}&= 1 - 2t^{2} \text{Re}(\ol b a) + t^{4}|a|^{2}|b|^{2} \\&= t^{2}(1 - 2\text{Re}(\ol b a) + |a|^{2}|b|^{2}) + (1 - t^{2})(1 - t^{2}|a|^{2}|b|^{2}) \\&\ge t^{2}(1 - 2\text{Re}(\ol b a) + |a|^{2}|b|^{2}) = t^{2}|1 - \ol ba|^{2}\end{align*}

\begin{align*}|\varphi_{t}(ta)|^{2}(|ta - tb|^{2} + |tc|^{2})&= t^{2}\abs{\frac{ta - tb}{1 - t^{2}\ol b a}}^{2}(|a - b|^{2} + |c|^{2}) \\&\le t^{2} \abs{\frac{a - b}{1 - \ol b a}}^{2}(|a - b|^{2} + |c|^{2}) \\&\le |ta - tb|^{2}\end{align*}

and $\norm{tT}_{M_2(\complex)}\le 1$ for all $t \in (0, 1)$. Therefore $\norm{T}_{M_2(\complex)}\le 1$.$\square$

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