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Proof of Isometry

  • Thread starter CrazyIvan
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1. Homework Statement
Prove or give a counterexample: if [tex]\mathcal{S} \in \mathcal{L} \left( V \right) [/tex] and there exists and orthonormal basis [tex]\left( e_{1} , \ldots , e_{n} \right)[/tex] of [tex]V[/tex] such that [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex] for each [tex]e_{j}[/tex], then [tex]\mathcal{S}[/tex] is an isometry.


2. Homework Equations
[tex] \mathcal{S} [/tex] is an isometry if
[tex]\left\| \mathcal{S} v \right\| = \left\| v \right\| [/tex]
for all [tex]v \in V[/tex].


3. The Attempt at a Solution
I figure that if I can prove that [tex]\mathcal{S}[/tex] is normal, then I can use the Spectral Theorem to prove that all every basis vector is an eigenvector.

Then I can prove that the square of the absolute value of the eigenvalue for each eigenvector equals 1, and the proof comes easy after that.

But I'm getting hung up on proving if it is normal.

Thanks in advance for any help.
 

Answers and Replies

HallsofIvy
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You don't need eigenvalues. Write v in terms of the basis: [itex]\vec{v}= a_1\vec{e_1}+ a_2\vec{e_2}+ \cdot\cdot\cdot+ a_n\vec{e_n}[/itex]. What is its length? What is Lv? What is its length?

You say [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex]. I would have interpreted [tex]\mathcal{S} \in \mathcal{L} \left( V \right) [/tex] to mean "linear transformations from V to itself". Do you mean the dual space of V: linear transformations from V to R?
 
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You don't need eigenvalues. Write v in terms of the basis: [itex]\vec{v}= a_1\vec{e_1}+ a_2\vec{e_2}+ \cdot\cdot\cdot+ a_n\vec{e_n}[/itex]. What is its length? What is Lv? What is its length?

You say [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex]. I would have interpreted [tex]\mathcal{S} \in \mathcal{L} \left( V \right) [/tex] to mean "linear transformations from V to itself". Do you mean the dual space of V: linear transformations from V to R?
Sorry for my lack of clarity. [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex] is a given property of the linear transformation [tex] \mathcal{S} [/tex]. So the only thing I know about [tex] \mathcal{S} [/tex] is that it's a linear transformation and it doesn't change the length of the orthonormal basis vectors.


I think I have a solution, but I'm not sure if it's the same thing you were saying:

[tex] \left\| S e_{i} \right\| = \left\| e_{i} \right\| [/tex]
[tex]\left\langle S e_{i} , S e_{i} \right\rangle = \left\langle e_{i} , e_{i}
\right\rangle[/tex]
[tex] \left\langle e_{i} , S^{*} S e_{i} \right\rangle = \left\langle e_{i} , e_{i} \right\rangle [/tex]
[tex] \left\langle e_{i} , S^{*} S e_{i} - e_{i} \right\rangle = 0 [/tex]
So since [tex] e_{i} [/tex] is not the zero vector,
[tex]S^{*} S e_{i} - e_{i} = 0 \ \Leftrightarrow S^{*} S e_{i} = e_{i} [/tex]
And from there, it's easy.

Does that look right?
 
Dick
Science Advisor
Homework Helper
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Sorry for my lack of clarity. [tex]\left\| \mathcal{S} e_{j} \right\| = 1 [/tex] is a given property of the linear transformation [tex] \mathcal{S} [/tex]. So the only thing I know about [tex] \mathcal{S} [/tex] is that it's a linear transformation and it doesn't change the length of the orthonormal basis vectors.


I think I have a solution, but I'm not sure if it's the same thing you were saying:

[tex] \left\| S e_{i} \right\| = \left\| e_{i} \right\| [/tex]
[tex]\left\langle S e_{i} , S e_{i} \right\rangle = \left\langle e_{i} , e_{i}
\right\rangle[/tex]
[tex] \left\langle e_{i} , S^{*} S e_{i} \right\rangle = \left\langle e_{i} , e_{i} \right\rangle [/tex]
[tex] \left\langle e_{i} , S^{*} S e_{i} - e_{i} \right\rangle = 0 [/tex]
So since [tex] e_{i} [/tex] is not the zero vector,
[tex]S^{*} S e_{i} - e_{i} = 0 \ \Leftrightarrow S^{*} S e_{i} = e_{i} [/tex]
And from there, it's easy.

Does that look right?
That doesn't prove (S*)S(e_i)-e_i=0. It just shows (S*)S(e_i)-e_i is orthogonal to e_i. I would suggest you take some time out from trying to prove it and put a little more energy into looking for a counterexample.
 

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