Orthogonal Projection

Question

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!

Answer 1

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

Answer 2

Here is an approach which doesn't use the decomposition of $V$ into a subspace and its orthogonal complement, nor finite dimension for that matter. However, I'll use the following characterization of orthogonal projection: A linear map $T\colon V\to V$ such that $T^2=T$ and such that $\langle v-T(v),T(w)\rangle=0$ for all $v,w$.

I'll take the complex case as there are some extra details, but the real case is analogous.

So we use the inequality $\Vert T(a)\Vert^2\leq\Vert a\Vert^2$ with $a=v+T(t w)$, where $v,w\in V$ and $t>0$: \begin{align*}\Vert T(v+T(tw))\Vert^2&\leq\Vert v+T(t w)\Vert^2\\\Vert T(v)\Vert^2+2\operatorname{Re}\langle T(v),T(t w)\rangle+\Vert T(t w)\Vert^2&\leq\Vert v\Vert^2+2\operatorname{Re}\langle v,T(t w)\rangle+\Vert T(t w)\Vert^2\\\Vert T(v)\Vert^2+2\operatorname{Re}\langle T(v),T(t w)\rangle&\leq\Vert v\Vert^2+2\operatorname{Re}\langle v,T(t w)\rangle\\t^{-1}\Vert T(v)\Vert^2+2\operatorname{Re}\langle T(v),T(w)\rangle&\leq t^{-1}\Vert v\Vert^2+2\operatorname{Re}\langle v,T(w)\rangle,\end{align*} where the last line is obtained from the previous one because $t>0$. Letting $t\to\infty$ we obtain \begin{align*}2\operatorname{Re}\langle T(v),T(w)\rangle\leq 2\operatorname{Re}\langle v,T(w)\rangle\end{align*} for all $v,w$. Simplifying the "$2$" term and using the same inequality with $-v$ in place of $v$ yields $$\operatorname{Re}\langle T(v),T(w)\rangle=\operatorname{Re}\langle v,T(w)\rangle$$ for all $v,w$. For a complex number $z$, we have $\operatorname{Im}(z)=\operatorname{Re}(-iz)$, so using the equality above with $iw$ in place of $w$ yields $$\operatorname{Im}\langle T(v),T(w)\rangle=\operatorname{Im}\langle v,T(w)\rangle$$ for all $v,w$. So the last two equalities give us $$\langle T(v),T(w)\rangle=\langle v,T(w)\rangle$$ for all $v,w$, and this means that $T$ is an orthogonal projection (onto $T(V)$), according to the definition above.