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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!

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I think you want "$\leq$" in your first line? –  Martin Argerami Mar 28 '12 at 5:51
    
You're right. I just didn't include the < sign because I was curious if the equality immediately implied that T was an orthogonal projection. I'm still not clear if it does, actually. Does orthogonality and then the fact that T is a projection, make it an orthogonal projection, or are those two things not related at all? –  Chris Mar 28 '12 at 21:21
    
Chris, if $T$ is an idempotent and it is isometric, then it has to be the identity (because the identity is the only idempotent with no kernel). –  Martin Argerami Mar 28 '12 at 22:07
    
Makes sense. Thanks! –  Chris Mar 29 '12 at 2:40

1 Answer 1

up vote 1 down vote accepted

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 $\Re iz = -\Im z$.

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Thank you so much for this; I completely understood the first part, but some part of the converse escaped me. For the last part, did you just show that by a choice of Lamba, you can make the real and complex parts of < x, Ty > both 0? Also, I don't really know what the "J" looking symbol means, so you please enlighten me? Thanks. –  Chris Mar 29 '12 at 2:50
    
That last sentence is supposed to read: "...so could you please enlighten me?". Bad typo. –  Chris Mar 29 '12 at 5:13
    
@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. –  martini Mar 29 '12 at 12:26

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