$ Av=\lambda v \implies\ A^*v=\overline \lambda v $, Is this true only for normal operators?

Solution 1:

Normality is essential. Note that $$\begin{align*}(A^*-\overline{\lambda})^*(A^*-\overline{\lambda})&=(A-\lambda)(A^*-\overline{\lambda})\\&=AA^*-\lambda A^*-\overline{\lambda}A+|\lambda|^2\\ &=A^*A-\lambda A^*-\overline{\lambda}A+|\lambda|^2\\ &=(A^*-\overline{\lambda})(A-\lambda)\\&=(A-\lambda)^*(A-\lambda). \end{align*}$$ This shows that if $A$ is normal, then $A-\lambda$ is also normal. Since $\|Nx\|=\|N^*x\|$ for every normal operator $N$, it also holds that $\ker N = \ker N^*$. Thus $$ \ker (A-\lambda) = \ker(A^*-\overline{\lambda}) $$ and it follows that $$ Av=\lambda v\ \ \ \Longleftrightarrow \ \ \ A^*v = \overline{\lambda}v. $$

EDIT (Necessity of normality, 1/3/2022)

To see that the normality of $A:V \to V$ is essential, we proceed by induction on the dimension of $V$.

If $\operatorname{dim}V = 1$, $A$ being normal is trivial.

Suppose $\operatorname {dim} V >1$. By the algebraic closedness of $\mathbb C$, we can find $w \ne 0$ and $\lambda \in \mathbb C$ such that $A w = \lambda w$. By the given condition, it is straightforward to see that $A^*w = \overline{\lambda} w$, and accordingly, $\langle w\rangle^\perp \equiv \left \{ v\in V: \langle v, w\rangle = 0\right\}$ is an $A$-invariant subspace. Now by the inductive hypothesis, $B \equiv A|_{\langle w\rangle^\perp} : \langle w\rangle^\perp \to \langle w\rangle^\perp$ is normal, which implies the normality of $$ A = \left[ \begin{matrix} \lambda & 0 \\ 0 & B \end{matrix} \right]. $$