TITLE: Orthogonal Projection
QUESTION [11 upvotes]: 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!

REPLY [2 votes]: 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.