Orthogonal Projectors and Operators If we have a projection T on a finite-dimensional inner product V, and we know that ||Tx|| = ||x||, can we conclude that T is an orthogonal projection?
The equality with the norms is enough to guarantee orthogonality of T, so I guess what I'm asking is: what is the difference between an orthogonal operator and an orthogonal projection?
On the same subject, how would a proof of the converse go? If we have an orthogonal projection  T, how would we prove that ||Tx|| <= ||x||?
My idea for this was to start with 0 <= ||Tx||^2 = < Tx, Tx > = < x, Tx > and then look at this last inner product in terms of an orthonormal basis of eigenvectors of T (we know this exists since T is normal). I got stuck at this point though; I couldn't argue that the equality becomes an inequality after you pull out all the eigenvalues. Suggestions?
Any help is appreciated, thanks!
 A: if $T$ is orthogonal, then 
$$
  \|Tx\|^2 = (Tx, Tx) = (x, Tx) \le \|x\|\|Tx\|
$$ 
by Cauchy-Schwarz, hence $\|Tx\| \le \|x\|$.
Now let $T$ be a projection with $\|Tx\| \le \|x\|$. Let $x \in \ker T$, $\lambda \in \mathbb C$ and $y \in X$. We have $T(x + \lambda Ty) = Ty$, hence
$$
  \|\lambda Ty\|^2 \le \|x + \lambda y\|^2 = \|x\|^2 + 2\Re \bar\lambda(x, Ty) + \|\lambda Ty\|^2.
$$
Hence $-2\Re\bar\lambda (x,Ty) \le \|x\|^2$. As $\lambda \in \mathbb C$ was arbitrary, we get $\Re (x, Ty) = 0$ by choosing $\lambda \in \mathbb R$ and $\Im (x, Ty) = 0$ by choosing $\lambda \in i\mathbb R$. Therefore $(x, Ty) = 0$ and $T$ is orthogonal.

To be more concrete in showing $(x, Ty)= 0$ in the last paragraph above: If we let $\lambda = -\frac 12\mu\Re(x,Ty)$ for $\mu \in \mathbb R$, we get 
$$
  \|x\|^2 \ge -2\Re\lambda(x,Ty) = \mu|\Re(x,Ty)|^2 \iff
    |\Re(x,Ty)|^2 \le \frac 1\mu \|x\|^2.
$$
Letting $\mu\to\infty$ gives $\Re(x,Ty) = 0$.
Now we choose $\lambda = \frac i2 \mu\Im(x,Ty)$. Now 
$$
  \Re\bar\lambda(x,Ty) = \frac 12\mu\Im(x,Ty) \Re i(x,Ty) 
                       = -\frac 12\mu|\Im(x,Ty)|^2
$$
and as above it follows that $\Im(x,Ty) = 0$. For the last equality we used $\Re iz = -\Im z$.
