Norm of Adjoint

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Theorem

Let $H, K$ be Hilbert spaces.

Let $A \in \map B {H, K}$ be a bounded linear transformation.

Let $A^* \in \map B {K, H}$ be the adjoint of $A$.


Then $A$ and $A^*$ satisfy:

$\norm A_{\map B {H, K} }^2 = \norm {A^*}_{\map B {K, H} }^2 = \norm {A^* A}_{\map B {H, H} }$

where:

$\norm \cdot_{\map B {H, K} }$ denotes the operator norm on $\map B {H, K}$
$\norm \cdot_{\map B {K, H} }$ denotes the operator norm on $\map B {K, H}$
$\norm \cdot_{\map B {H, H} }$ denotes the operator norm on $\map B {H, H}$


Proof

Let $h \in H$ such that $\norm h_H \le 1$.

Then:

\(\ds \norm {A h}_K^2\) \(=\) \(\ds \innerprod {A h} {A h}_K\) Definition of Inner Product Norm
\(\ds \) \(=\) \(\ds \innerprod {A^* A h} h_H\) Definition of Adjoint Linear Transformation
\(\ds \) \(\le\) \(\ds \norm {A^*A h}_H \norm h_H\) Cauchy-Bunyakovsky-Schwarz Inequality
\(\ds \) \(\le\) \(\ds \norm {A^* A} \norm h_H^2\) Fundamental Property of Norm on Bounded Linear Transformation
\(\ds \) \(\le\) \(\ds \norm {A^* A}\) Assumption on $h$
\(\ds \) \(\le\) \(\ds \norm {A^*} \norm A\) Norm on Bounded Linear Transformation is Submultiplicative

By definition $(1)$ for $\norm A$, it follows that:

$\norm A^2 \le \norm {A^* A} \le \norm {A^*} \norm A$

That is:

$\norm A \le \norm {A^*}$.


By substituting $A^*$ for $A$, and using $A^{**} = A$ from Adjoint is Involutive, the reverse inequality is obtained.


Hence $\norm A^2 = \norm {A^* A} = \norm {A^*}^2$.

$\blacksquare$


Sources

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