1
$\begingroup$

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!

  • 0
    Makes sense. Thanks!2012-03-29

1 Answers 1

1

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 $\Re iz = -\Im z$.

  • 0
    @Chris: Your are right. $\Re z$ denotes the real part of $z\in \mathbb C$, $\Im z$ its imaginary part. I will add something about the choice of $\lambda$ above.2012-03-29