# Orthogonal Projection

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!

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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

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