Kernel of Normal Operator is Kernel of Adjoint

Theorem

Let $H$ be a Hilbert space.

Let $A \in \map B H$ be a normal operator.

Then:

$\ker A = \ker A^*$

where:

$\ker$ denotes kernel
$A^*$ denotes the adjoint of $A$.

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:

$\ker A = \ker A^*$

$\blacksquare$