# Question about a proof.

1. Jul 11, 2009

### evilpostingmong

Suppose T ∈ L(V) and U is a subspace of V. Prove that U is
invariant under T if and only if U⊥ is invariant under T∗.

Now for reference, L(V) is the set of transformations that map v (a vector) from V to V.
T* is the adjoint operator.
The case where the dimension of U is less than V bugs me. How can U
be invariant under T* when T* maps from U to V unless mapping to V
also counts as mapping to U since U is in V itself. Now when
I say map to V, I mean lets say V is 3 dimensional and U is 1 dimensional.
Then u=(x3) gets mapped (by T*) to a vector with three nonzero components.
Does this count as mapping from U to U?
I'm just a bit confused about this, and any help will be greatly appreciated.
I guess that I'm confused enough that my post doesn't make much sense,
so in that case, could someone nudge me in the right direction (ie give a good hint)?
Thank you!

Last edited: Jul 11, 2009
2. Jul 12, 2009

### evilpostingmong

Well, ok the thing that I don't understand is how nullT* can even
exist if T* sends a vector from a space of lower dimension to a space of higher dimension
unless 0 is the only element in nullT*. So if T* maps a vector from 2d space to 3d space,
what is within nullT*? Sorry for double posting, but I need to know badly. Thank you!

3. Jul 12, 2009

### Office_Shredder

Staff Emeritus
Umm... T* sends a vector from V to V. It's defined to be the linear transformation such that <Tu,v> = <u,T*v>

The definition of invariance is U is invariant under T if T(u) is in U for all u in U. So we want to show T maps elements of U to U if and only if T* maps elements of U to U. Start with the definition of T* above and the definitino of U

4. Jul 13, 2009

### evilpostingmong

Let v be a vector in V, let u be a vector in U and let w be a vector in U$$\bot$$. Let v=u+w. Now <Tu, u>+<Tu, w>=<u, T*u>+<u, T*w>
If T*w maps from U$$\bot$$ to U$$\bot$$, then <u, T*w>=0 therefore
<Tu, u>+<Tu, w>=<u, T*u> as a result. Now given that <Tu, u>=<u, T*u> (apply * to both sides gives
<T*u, u>+<u, T**u>=<T*u, u>+<u, Tu>) this equation holds when w is orthogonal to Tu which forces
T to map u from U to U as a result of w being orthogonal to Tu and to make <Tu, u>+<Tu, w>=<u, T*u> true.
Note that we wouldn't have arrived at <Tu, u>+<Tu, w>=<u, T*u> if T* didn't map w from U$$\bot$$
to U$$\bot$$ and this equation allows <Tu, w> to=0 which allows Tu to map from U to U
since all elements u in U are orthogonal to all elements w in U$$\bot$$.

Last edited: Jul 13, 2009
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