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Let T be a normal transformation, $ T : \mathbb{C}^n \to \mathbb{C}^n $, and let $ 0 \neq v \in \mathbb{C}^n $. Prove that if $ \{Sp(v)\}^{\perp} $ is $T$-invariant, then $ v $ is an eigenvector of $ T$. One possible clue: Is $ T^{*}(v) \perp \{Sp(v)\}^{\perp} $ true? My guess is that because $T$ is normal then $ ImT = (ImT)^* $, but I'm not sure how to follow from there.

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If $T^*(v)$ is orthogonal to $v$, then you would have $\langle T(v),v)\rangle =\langle v,T^*(v)\rangle=0$ which would mean that $T(v)$ is orthogonal to $v$. Then $v$ could not be an eigenvector for $T$. –  Joe Johnson 126 Nov 13 '11 at 22:49
    
EDIT: The clue is an anti-clue :) –  dan Nov 13 '11 at 22:59
    
EDIT2: I've had a typo in the clue, thanks for spotting it :) –  dan Nov 13 '11 at 23:01

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up vote 2 down vote accepted

Your updated clue is the same as asking if $v$ is an eigenvector of $T^*$, which it is: If $x$ is orthogonal to $v$ then

$$ \langle x,T^*(v)\rangle = \langle T(x),v\rangle =0 $$

since $\operatorname{Span}(v)^\perp$ is $T$-invariant. But that means that $T^*(v)$ is in $\left(\operatorname{Span}(v)^\perp\right)^\perp=\operatorname{Span}(v)$. Since $T$ and $T^*$ have the same eigenvectors, you're done.

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