I know that if an operator $T$ in $L(V)$ (where $V$ is a finite dimentional vector space over the complex field) is normal than 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?

• Counterexample $\mathrm{span}\{(1,0)\}\subset \mathbb{R}^2$ is invariant subspace of $$T=\begin{Vmatrix} 1& 1\\0 &1\end{Vmatrix},$$ but not invariant subspace of $T^*$. – Norbert Jun 29 '12 at 13:57
• You need $V$ to have an inner product to talk about adjoints, and taking adjoints doesn't preserve invariant subspaces in general. (It does if $T$ is normal, but this just reflects the general fact that commuting operators preserve each other's eigenspaces). – Qiaochu Yuan Jun 29 '12 at 13:59

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 $\begin{bmatrix}0\\1\end{bmatrix}$.

The problem of your reasoning is that $\langle v,\lambda^*w\rangle=\langle \lambda v,w\rangle=\langle Tv,w\rangle=\langle v,T^*w\rangle$ does not imply $T^*w=\lambda^*w$. For this to be true, you must have $\forall v \in V, \langle \lambda v,w \rangle = \langle v,T^*w \rangle$. However, this equality only holds for eigenvector one specific eigenvector of $T$ under this circumstance.

For the proof you need:

(1) if $T$ is normal than $kerT = kerT^*$

Take any $v\in$$V than because T is normal you get$$<T(v)|T(v)> = <T^*(v)|T^*(v)>$$So every v\inkerT iff also \in$$kerT^*$

(2) if T is normal than $T-$$\lambda$$Id$ is normal. (this is easy to check).

Now if $v$ is an eigenvector of $T$ with eigenvalue $\lambda$, than $v$$\in$$ker(T-$$\lambda$$Id)$.

Now from (2) we get that $T-$$\lambda$$Id$ is normal, so from (1) we get that $ker(T-$$\lambda$$Id)$ = $ker(T^*-$$\overline\lambda$$Id)$ so than we can say that $v$ is an eigenvector of $T^*$ with eigenvalue of $\overline\lambda$.