Prove the following: If $T \in \mathcal{L}(V)$, and $v$ is an eigenvector of $T$, then $\overline{v}$ is an eigenvector of $T^*$.

Question

I feel like this should be obvious, but I'm kind of stuck on how to prove it.

I know that $v \in \text{null}(T - \lambda I)$, and that $\overline{\lambda}$ is an eigenvalue of $T^*$, but I don't really know how to use that information to prove this. Any help would be appreciated

(Btw, you can assume $V$ is a f.d.v.s. over $\mathbf{C}$.)

Answer 1

This doesn't hold. Consider the transformation $T \in \mathcal L(\mathbb C^2)$ whose matrix representation is $$ A = \left[ \begin{array}{cc} 1&4i \\ i&1 \end{array} \right].$$ Then the vector $(2,1)$ is an eigenvector of $A$ with eigenvalue $1+2i$ (check this!).

On the other hand, $\bar{v}$ is not an eigenvector of $A^*$. Clearly, we have $\bar{v} = (2,1)$ and $$A^* = \left[\begin{array}{cc} 1&-i \\ -4i& 1 \end{array} \right],$$ and so we see that $$\left[\begin{array}{cc} 1&-i \\ -4i& 1 \end{array} \right] \left[ \begin{array}{c} 2\\1 \end{array} \right] = \left[\begin{array}{c} 2-i \\ 1-8i \end{array} \right],$$ i.e. the vector $(2,1)$ is not an eigenvector of $A^*.$