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Proposition: Suppose that $ V $ is a complex vector space and $ \dim(V) < \infty $. Then $ T \in \mathcal{L}(V) $ is normal if and only if the orthogonal complement of every $ T $-invariant subspace is $ T $-invariant.

I hope that you can help me with a solution or a hint. Thanks.

My idea:

The forward implication: If $ T $ is normal, then $ T^{*} = p(T) $ for any polynomial $ p \in \Bbb{C}[X] $. Then given a $ T $-invariant subspace $ U $, we know that $ U $ is $ p(T) $-invariant. In other words, $ U $ is $ T^{*} $-invariant. As $ U $ is $ T^{*} $-invariant, it follows that $ W \stackrel{\text{df}}{=} U^{\perp} $ is $ (T^{*})^{*} $-invariant. Hence, $ W $ is $ T $-invariant.

I was unable to work out the backward implication.

Berrick Caleb Fillmore
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João
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3 Answers3

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Note that $T$ will have at least one eigenvector since the ground field is algebraically closed. If $v$ is a $T$ eigenvector and $W\perp \{v\}$ then for $w \in W$,

$$(T^*v,w)=(v,Tw)=0$$ so in fact $T^*v$ is orthogonal to $W$ and thus $v$ is an eigenvector of $T^*$ also. It is now easy to see that $W$ is $T^*$ invariant and the same assumptions about $T$ hold restricted to $W$, now by induction $T$ is normal restricted to $W$ and this gives that $T$ is normal.

Or better the assumptions imply that $T$ is diagonalizabe by the argument I gave.

Rene Schipperus
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Hint: Use for a normal operator $||Tv|| = ||T^*v||$. What does this tell you when you apply this on the matrix induced by $T$

MathMan
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  • I thought about it but did not get anywhere. – João Jul 05 '14 at 18:15
  • Any vector space $V = U + U'$ where $U'$ is the orthogonal complement of $U$. Now, $U$ is given to be invariant under a linear mapping $T$. Can you write the matrix induced by $T$ now.

    Now use that $||Tv||=||T^*v||$. Can you visualize something?

     – MathMan Jul 05 '14 at 18:20
  • What will the product of the matrices induced by $T$ and $T^*$ look like? – MathMan Jul 05 '14 at 18:27
  • Sorry, I can't see it. – João Jul 05 '14 at 18:28
  • You should also take into account that for an orthonormal basis, the matrix induced by $T$ and $T*$ are the conjugate transposes of each other. you might want to look up Linear Algebra done right by sheldon axler. – MathMan Jul 05 '14 at 18:31
  • @VHP Your hint seems to be relevant only for the already proved implication (if $T$ is normal, then ...). The question is about the converse. – Etienne Jul 05 '14 at 18:39
  • @Etienne for an orthonormal basis, matrix induced by $T$ and $T^*$ are conjugate transposes of each other, and any $v=U \oplus U'$. I believe this should give the desired result? – MathMan Jul 05 '14 at 18:52
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    yess, but I like to make sure it was right .. – João Jul 05 '14 at 18:52
  • I hope you get it :-) – MathMan Jul 05 '14 at 18:57
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    @VHP Read again your hint: "for a normal operator ...". And also your first comment: "Now use that $\Vert Tv\Vert=\Vert T^*v\Vert$. Here, you don't know that $T$ is normal; this is what you want to prove. – Etienne Jul 06 '14 at 08:53
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It is not true that $T^*=p(T)$ for any polynomial in $\mathbb{C}(X)$. However, there exists polynomials $p$ in $\mathbb{C}(X)$ such that $p(T)=T^*$. It suffices to take a polynomial which maps any eigenvalue $\lambda\in\mathbb{C}$ in $\bar{\lambda}$. The interpolation polinomial will do that...

user623860
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