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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.


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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
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 $Ker T^\bot \subset Im T$.

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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

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