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.


Solutions Collecting From Web of "Orthogonal Projection"

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