Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Seems like I still don't get it, I think I am missing something important.

Let $V$ be an $n$ dimensional inner product space ($n \geq 1$), and $T\colon\mathbf{V}\to\mathbf{V}$ be a linear transformation such that:

  • $T^2 = T$
  • $||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;

Prove that a subspace $U \subseteq V$ exist, such that $T$ is the orthogonal projection on $U$.

Now, I know these things:

  • The fact hat $T^2 = T$ guarantees that $T$ is indeed a projection, so I need to prove that T is an orthogonal projection (I guess this is where $||T(a)|| \leq ||a||$ kicks in).
  • To do this I can prove that:
    • For every $v$ in $ImT^{\perp}$, $T(v) = 0$
    • Alternatively, I can prove that for every $v$ in $ImT$ and $u$ in $KerT$, $(v,u)=0$.
    • $T$ is self-adjoint (according to Wikipedia)
    • The matrix $A = [T]_{E}$ when $E$ is an orthonormal basis, is hermitian (this is equivalent to the previous point).
    • What else?

I've been thinking about it for quite some time now, and I'm pretty sure there is something big I'm missing, again. I just don't know how to use the data to prove any of these things.

Thanks!

share|improve this question
4  
Rule of thumb: if you don't know how to prove something, try to find a counterexample. Eventually you will say to yourself "well, of course I can't find a counterexample, because (blank) is stopping me," and you can probably turn (blank) into a proof. –  Qiaochu Yuan Feb 12 '11 at 15:13
add comment

1 Answer

up vote 3 down vote accepted

One approach to this is by using Pythagorean Theorem. You fill in the details.

Indeed, suppose that $T \psi \in Im T$. Note that $V = Ker T \oplus Ker T^\bot$. Now write $T \psi = k + k'$ where $k \in Ker T$ and $k' \in Ker T^\bot$. We get that

$$\|k'\|^2 \ge \|T k'\|^2 = \|T k + T k'\|^2 = \|T(T \psi)\|^2 = \|T \psi\|^2 = \|k\|^2 + \|k'^2\|,$$

from which we gather that $k = 0$. Therefore $T \psi \in Ker T^\bot$. Thus $Im T \subset Ker T^\bot$. On the other hand if $\phi \in Im T^\bot \cap Ker T^\bot$, then one can show that $\phi \in Im T^* \cap Ker T^*$. (Because $Im T^\bot = Ker T^*$.) One can also show that $T^*$ is a projection. Let $\phi = T^* \psi$. Then $T^* \psi = T^* T^* \psi = T^* \phi = 0$. Hence $\phi = 0$. Since $V = Im T \oplus Im T^\bot$, we get that $Ker T^\bot \subset Im T$.

share|improve this answer
    
Thanks! May I ask how you came up with the first part? It's easy for me to think about things like the second part of your proof (because I just "visualize" it), but what you did with the theorem just seems like symbol manipulation to me. I mean, I can see why it is working, now that it is laid before me, but I wouldn't know how to get there if I had to. –  Hila Feb 12 '11 at 21:36
    
Hila: Qiaochu's advice is sound and drawing a picture helps, too. If you draw a non-orthogonal projection onto a line in the plane along with its Im, Ker and Ker^\bot, you'll see quite quickly what's going to go wrong with the second condition you have. Your intuition here should be that if the projection is non-orthogonal then there will be a vector whose length gets increased in the mapping. Taking the direct sum with Ker T and Ker^\bot is also a quite natural thing to try, since that will allow you to concentrate separately on the part of the vector which is significant in the projection. –  J. J. Feb 13 '11 at 7:36
    
... continued ... And of course taking that direct sum lets you to use Pythagorean theorem as well, which is a good tool for calculating lengths of vectors. –  J. J. Feb 13 '11 at 7:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.