Let $x$ be an eigenvector of $A.$ Is it true that if $x^{\perp}$ is invariant under $A,$ then $A$ is normal? Let $A$ be an $n \times n$ matrix with entries in $\mathbb C,$ and let $x$ be an eigenvector of $A.$ If $x^{\perp}$ is invariant under A, is it true that $A$ is normal?
Here is my idea.

Let $\lambda$ be an eigenvalue of $A$ such that $Ax = \lambda x.$ Let $y \in x^{\perp}.$ Then, we have that $\overline{\lambda} y \in x^{\perp}$ and $A(\overline{\lambda} y) \in x^{\perp}$ so that $\left<A(\overline{\lambda}y),x\right>=\left<Ay,\lambda x\right>=\left<Ay,Ax\right>.$ So, $A$ is normal.

Thanks in advance for your help.

This is false. $x=(1,0,0)^T$ is an eigenvector of $A=\pmatrix{0&0&0\\ 0&0&1\\ 0&0&0}$ and $x^\perp$ is $A$-invariant, but $A$ isn't normal.
The statement can be corrected by requiring that $x^\perp$ is $A$-invariant for every eigenvector $x$ of $A$. The corrected statement can be proved by mathematical induction on the size of $A$ or by constructing an orthogonal eigenbasis of $A$.
