Kernel of Normal Operator is Kernel of Adjoint
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Theorem
Let $H$ be a Hilbert space.
Let $A \in \map B H$ be a normal operator.
Then:
- $\ker A = \ker A^*$
where:
Proof
Let $x \in H$ be arbitrary.
Then:
\(\ds x\) | \(\in\) | \(\ds \ker A\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds A x\) | \(=\) | \(\ds \mathbf 0_H\) | Definition of Kernel of Linear Transformation | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \gen {A x, A x}\) | \(=\) | \(\ds 0\) | Definition of Inner Product | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \gen {A^*Ax, x}\) | \(=\) | \(\ds 0\) | Definition of Adjoint Linear Transformation | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \gen {A A^* x, x}\) | \(=\) | \(\ds 0\) | Definition of Normal Operator | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \gen {x, AA^*x}\) | \(=\) | \(\ds 0\) | Definition of Inner Product | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \gen {A^*x, A^* x}\) | \(=\) | \(\ds 0\) | Definition of Adjoint Linear Transformation | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds A^*x\) | \(=\) | \(\ds \mathbf 0_H\) | Definition of Inner Product | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds \ker A^*\) | Definition of Kernel of Linear Transformation |
Hence, by definition of set equality:
- $\ker A = \ker A^*$
$\blacksquare$