I know that if an operator $T$ in $L(V)$ (where $V$ is a finite dimentional vector space over the complex field) is normal, then for every vector $v$ in $V$ , $Tv= \lambda v$ iff $T^*v= \lambda^*v$ (where $\lambda^*$ is the complex conjugate of $\lambda$). Why isn't this correct for every operator?
I think that you can take the eigenspace of $\lambda$ (regarding $T$) and this is a $T$-invariant subspace of $V$. So this is also a $T^*$ -invariant subspace and there you get that for every $v,w$ in $\mathrm{eigenspace}(\lambda)$ $\langle v,\lambda^*w\rangle=\langle \lambda v,w\rangle=\langle Tv,w\rangle=\langle v,T^*w\rangle$ and therefor $T^*w=\lambda^*w$. What is my mistake?
Best Answer
Your mistake is assuming that if $\lambda$ is eigenvalue of $T$ with eigenvector $v$, then $\overline\lambda$ is eigenvalue of $T^*$ (this is true) also with eigenvector $v$ (this is not true in general; it is when $T$ is normal).
Using Norbert's example, $1$ is eigenvalue of $T$ with eigenvector $v=\begin{bmatrix}1\\0\end{bmatrix}$. But $v$ is not an eigenvector of $T^*$: $$ T^*v=\begin{bmatrix}1&0\\1&1\end{bmatrix}\,\begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}1\\1\end{bmatrix} $$ Still, of course, $1$ is indeed an eigenvalue of $T^*$, but with eigenvector $\begin{bmatrix}0\\1\end{bmatrix}$.