# Isometries math problem

1. Aug 14, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
Prove or give a counterexample: if S ∈ L(V) and there exists
an orthonormal basis (e1, . . . , en) of V such that llSejll = 1 for
each ej , then S is an isometry.

2. Relevant equations

3. The attempt at a solution
Can't think of a counterexample. I am assuming that since ej is a member
of an orthonormal basis, llejll HAS to be 1 by definition. And if llSejll=1,
then llSejll=llejll. I mean, considering that llSejll=1 for ALL ej,
all ej have 1 as an eigenvalue. Should I go ahead and do the proof, because
I am having trouble seeing how S is not an isometry given the information I posted
above.

2. Aug 14, 2009

### GPPaille

Re: isometries

You could start by proving that every basis vector has the same length after transformation ( Hint: squared length ). And if every vector is a linear combination of these basis vectors, do the same proof has above, with a little more coefficients.

3. Aug 14, 2009

### Dick

Re: isometries

||Sej||=||ej||=1 DOES NOT mean 1 is an eigenvalue with ej as an eigenvector. IF ej is an eigenvector of S with eigenvalue 1, THEN ||Sej||=||ej||=1. The converse is emphatically NOT TRUE. You make this mistake A LOT. Stop it! Construct a counterexample in R^2. Quick. DON'T try to prove it. I know you will throw a 'proof' together. That would be a bad thing.

Last edited: Aug 14, 2009
4. Aug 15, 2009

### evilpostingmong

Re: isometries

I digress, but to clear up any misconceptions I have about isometries,
lets say we have a vector (2, 2). apply T (2x2 matrix of rank 1 with sqrt(8)/2 at
the far upper left corner and 0 everywhere else) to (2, 2) to get (sqrt(8), 0).
(2, 2) is not an eigenvector of T and T(2, 2)=/=1*(2, 2). But llT(2, 2)ll=ll(2, 2)ll.

(sqrt(2), sqrt(-1)) Let T be a 2x2 matrix of rank 1 with sqrt(1/2)at the far upper left corner and 0 everywhere else. T(sqrt(2), sqrt(-1))=(1, 0). Now applying T again we get
(sqrt(1/2), 0). llT^2(sqrt(2), sqrt(-1))ll=sqrt(1/2)=/=1. This shows that llTekll=llekll does not imply that
T is an isometry since all isometries are such that llT^kekll=llekll where k is any number >=1.

Last edited: Aug 15, 2009
5. Aug 15, 2009

### GPPaille

Re: isometries

$$\left\|(\sqrt{2},i)\right\| = (\sqrt{2},i) \cdot (\sqrt{2},i)^* = 3$$

6. Aug 15, 2009

### evilpostingmong

Re: isometries

Oh right. :rofl: Ok here's plan B. (sqrt(2/3), sqrt(1/3) ). Now
let T's 2x2 matrix be of rank 1 with sqrt(3/2) at the top left hand corner
and 0 everywhere else. Now T(sqrt(2/3), sqrt(1/3))=(1, 0). Now applying T
again, we get (sqrt(2/3), 0). ll(sqrt(2/3), 0)ll=sqrt(2/3).

7. Aug 15, 2009

### GPPaille

Re: isometries

But you don't have to apply T twice to see that (1,0) do not preserves length. Write it in term of the basis vectors. And you will see that only one basis vector preserves its length.