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I found the following question:

Let $V$ be a finite dimensional inner product space over $C$ and let $T$ in $A(V)$ be unitary transformation. Prove that if $U$ is a subset of $V$ and it is $T$-Invariant then also the orthogonal complement of $U$ is $T$-Invariant.

What does the set $A(V)$ means?

I tried to do the following: Let $x$ be in the orthogonal complement of $U$, we need to show that for every y in $U$ $(T(x),y) = 0$. If $T = T^*$ then $(T(x),y) = (x,T^*(y)) = (x,T(y)) = 0$ but in this case $T^* = T-1$ so $(T(x),y) = (x,T-1(y))$ but I don't know how to continue from here.

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  • $\begingroup$ I guess $\,A(V)\;$ is the set of all endomorphisms = all linear operators of $\,V\,$ to itself. BTW, you made almost all to prove what you want... $\endgroup$
    – DonAntonio
    Aug 5, 2013 at 17:00

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No idea what $A(V)$ means, but if $U$ is $T$-invariant (i.e. $T(u) \in U$ for all $u \in U$), then $U^{\top} = \{ w \in V \, | \, \forall u \in U, \quad (w,u) = 0 \}$ will also be $T$-invariant.

Since $T$ is unitary, there exists $S$ such that $T \circ S$ is the identity map on $V$, so that $T$ is an isomorphism of $V$ to itself. By $T$-invariance, the restriction of $T$ to $U$ is an isomorphism of $U$ to itself, because $T$ maps $U$ to $U$ and is injective, but being a linear map of finite-dimensional vector spaces it must also be surjective.

This being said, if you take $w \in U^{\top}$ and an arbitrary element of $U$, write it in the form $u = T(u')$, then $T(w)$ is also in $U^{\top}$ because $(T(w), T(u')) = (w,u') = 0$. Since $u = T(u')$ was arbitrary, we conclude $T(w) \in U^{\top}$ and $U^{\top}$ is $T$-invariant.

Hope that helps,

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  • $\begingroup$ How does the equality $\langle T(w), T(u') \rangle = \langle w, u \rangle$ follow? Also on the right-hand side there's a typo in your post: $u'$ should be $u$. $\endgroup$
    – Mussé Redi
    Mar 7, 2017 at 16:54
  • $\begingroup$ @Mussé Redi : There are different definitions of unitary which are all equivalent, and this is one of them. What is yours? By the way, I don't know of which right-hand side you are speaking of, but I read all my equations with a $u'$ in them and didn't see a typo. Maybe you understood me wrong somewhere? $\endgroup$
    – Patrick Da Silva
    Mar 7, 2017 at 17:04
  • $\begingroup$ My definition of unitary is the same as yours; I forgot. I see now why $u'$ is not a typo. $\endgroup$
    – Mussé Redi
    Mar 7, 2017 at 17:08
  • $\begingroup$ @Mussé Redi: don't try to be funny! I was serious, I don't see typos. $\endgroup$
    – Patrick Da Silva
    Mar 7, 2017 at 17:10
  • $\begingroup$ I thought it was a typo because I presumed you used $u=T(u')$ instead of the unitarity. $\endgroup$
    – Mussé Redi
    Mar 7, 2017 at 17:11
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Hints:

We're given that

$$\forall\,u\in U\;,\;\;Tu\in U$$

Take now any $\;w\in U^\perp\;$ , and let us denote by $\;\langle,\rangle\;$ the inner product in $\,V\,$ , so:

$$\forall\,u\in U\;,\;\;0=\langle Tu,w\rangle=\langle u,T^*w\rangle\implies T^*w\in U^\perp$$

since it was any $\,u\in U\;,\;w\in U^\perp\;$ , and this means $\,T^*U^\perp\subset U^\perp\;$

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  • $\begingroup$ He want to show $U^\perp$ is $T$-invariant, not $T^*$-invariant. $\endgroup$
    – Owen Sizemore
    Aug 5, 2013 at 19:15
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    $\begingroup$ @OwenSizemore, it's the same in this case...right? $\endgroup$
    – DonAntonio
    Aug 5, 2013 at 19:35
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    $\begingroup$ @DonAntonio Sorry if it seems trivial, but what makes them the same is what I was looking for. Please help me. $\endgroup$
    – S.S
    May 6, 2020 at 20:11

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