$A v = \lambda v \implies A^* v = \bar{\lambda} v$ if $A$ is normal [duplicate]

Question

I want to show that if $A$ is normal then

$$ A v = \lambda v \implies A^* v = \bar{\lambda} v $$

I can show that $A^*v$ is also an eigenvector of $A$, using the fact that $A$ and $A^*$ commute, but I know that this doesn't imply $A^* v \propto v$. So I'm not sure where to go from here. Thanks.

EDIT: The top answer solves the problem, but I would like to know how to prove this result using the polarisation identity as suggested below. This is not the approach taken in the other questions on this site, linked to in the comments below.

Answer 1

For a normal operator $A$, we have $\lVert Aw\rVert = \lVert A^\ast w\rVert$ for all $w$. Now, assuming that $v$ is an eigenvector to the eigenvalue $\lambda$ of $A$, we find

$$\langle A^\ast v, v\rangle = \langle v, Av\rangle = \langle v, \lambda v\rangle = \overline{\lambda}\langle v,v\rangle.\tag{1}$$

Further, we have

$$\lvert \langle A^\ast v, v\rangle\rvert \leqslant \lVert A^\ast v\rVert\cdot \lVert v\rVert = \lvert\lambda\rvert\cdot \lVert v\rVert^2\tag{2}$$

by the Cauchy-Bun'akovskij-Schwarz inequality. By $(1)$, we have the equality case, hence $A^\ast v$ and $v$ are linearly dependent, i.e. $A^\ast v = \mu\cdot v$ for some $\mu \in \mathbb{C}$. But then $(1)$ immediately yields $\mu = \overline{\lambda}$.

Answer 2

Or you do what Daniel did $\langle A^\ast v, v\rangle = \langle v, Av\rangle = \langle v, \lambda v\rangle = \overline{\lambda}\langle v,v\rangle.\tag{1}$ and by polarization you have what you want.