Let $A$ be a normal operator on $\mathbb C^n$. (Normal means $AA^\ast=A^\ast A$, where $A^\ast$ is the adjoint operator.) How do I show that $\ker A=(\operatorname{im} A)^\perp$? (Orthogonality is understood with respect to the standard complex dot product on $\mathbb C^n$.)
I cannot prove a single inclusion.
For example, let $x\in \ker A$. Let $y\in \operatorname{im} A$ so that $y=Az$. Need to prove $(x,y)=0$. $(x,y)=(x,Az)=(A^\ast x, z)$. But how to make use of the normality of $A$?
For $x\in(\text{Im}A)^{\perp}$, then for all $z$, $\left<A^{\ast}x,z\right>=\left<x,Az\right>=0$, so $A^{\ast}x=0$. Then $\left<Ax,Ax\right>=\left<x,A^{\ast}Ax\right>=\left<x,AA^{\ast}x\right>=0$, so $Ax=0$, this proves $(\text{Im}A)^{\perp}\subseteq\ker A$.