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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.

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  • $\begingroup$ I managed to make the first implication. $\endgroup$
    – João
    Jul 5, 2014 at 17:44
  • $\begingroup$ Great. Can you include your proof, in detail, in your post? $\endgroup$
    – Pedro ♦
    Jul 5, 2014 at 17:45
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    $\begingroup$ @ReneSchipperus If the OP is explicit about what he has already proven, he won't get answers telling him things he already knows. If he posts his attempts, he will get better hints based on his ideas, &c. $\endgroup$
    – Pedro ♦
    Jul 5, 2014 at 17:51
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    $\begingroup$ If $T$ is normal, then $T^\ast =p(T)$ for any polynomial...? $\endgroup$
    – Pedro ♦
    Jul 5, 2014 at 18:04
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    $\begingroup$ @João: Not true. The identity operator $ I $ is normal, but how can $ I = 2 I $? $\endgroup$
    – Berrick Caleb Fillmore
    Dec 17, 2015 at 23:06

3 Answers 3

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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.

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  • $\begingroup$ But I don´t know if T is diagonalizable or not. $\endgroup$
    – João
    Jul 5, 2014 at 18:27
  • $\begingroup$ What is your ground field ? $\endgroup$
    – Rene Schipperus
    Jul 5, 2014 at 18:28
  • $\begingroup$ complex field.. $\endgroup$
    – João
    Jul 5, 2014 at 18:32
  • $\begingroup$ I don't understand.. Can you be more especific? Because, we now assume that every complement ortoghonal of subspace T-invariante is T-invariant. Why this implies that T is normal? $\endgroup$
    – João
    Jul 5, 2014 at 18:46
  • $\begingroup$ @João Read again Rene's answer, especially the end of it. What he shows is that the assumption on $T$ implies that $T$ is diagonalizable in some orthonormal basis; which obviously implies that $T$ is normal. $\endgroup$
    – Etienne
    Jul 6, 2014 at 8:48
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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 $T$

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  • $\begingroup$ I thought about it but did not get anywhere. $\endgroup$
    – João
    Jul 5, 2014 at 18:15
  • $\begingroup$ 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? $\endgroup$
    – MathMan
    Jul 5, 2014 at 18:20
  • $\begingroup$ What will the product of the matrices induced by $T$ and $T^*$ look like? $\endgroup$
    – MathMan
    Jul 5, 2014 at 18:27
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    $\begingroup$ yess, but I like to make sure it was right .. $\endgroup$
    – João
    Jul 5, 2014 at 18:52
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    $\begingroup$ @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. $\endgroup$
    – Etienne
    Jul 6, 2014 at 8:53
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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...

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