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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?

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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
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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

1 Answer 1

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

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