- #1

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## Homework Statement

Let U be an mxn matrix with orthonormal columns, and let x and y be in R^n. Prove

(Ux).(Uy)=x.y

## Homework Equations

I know ||Ux||^2=(Ux)^T(Ux)

## The Attempt at a Solution

I just have no idea...

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- Thread starter Lonely Lemon
- Start date

- #1

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Let U be an mxn matrix with orthonormal columns, and let x and y be in R^n. Prove

(Ux).(Uy)=x.y

I know ||Ux||^2=(Ux)^T(Ux)

I just have no idea...

- #2

- 128

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Do you know anything about inverses of orthonormal matrices?

- #3

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- #4

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- #5

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(Ux).(Uy)=(x^T)(U^T).(y^T)(U^T)

It's like drawing blood from a stone - I get stuck at every step...

- #6

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(Ux)(Uy) = (x^T)(U^T)(Uy) = (x^T)[(U^T)(U)](y).

Now use what you know about (U^T)(U).

- #7

Dick

Science Advisor

Homework Helper

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Do you know anything about inverses of orthonormal matrices?

I don't think you can really do it that neatly. U is mxn, not nxn. It's not necessarily a square matrix. It doesn't necessarily have an inverse. Try working on the usual basis of R^n={e1,...,en}. U(e_i) is the ith column of U. You have to be a little more detail oriented here.

- #8

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I didn't realise you could break up the dot product components and just multiply them together

- #9

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I don't think you can really do it that neatly. U is mxn, not nxn. It's not necessarily a square matrix. It doesn't necessarily have an inverse. Try working on the usual basis of R^n={e1,...,en}. U(e_i) is the ith column of U. You have to be a little more detail oriented here.

Ah shiet. That is right.. I should have read more carefully.

To OP, What I have said so far works for orthonormal matrices but that is not what you have lol. Sorry.

- #10

Dick

Science Advisor

Homework Helper

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Ah shiet. That is right.. I should have read more carefully.

To OP, What I have said so far works for orthonormal matrices but that is not what you have lol. Sorry.

Carry on. It's easy to fix. U(e_i).U(e_j) is 1 if i=j and zero otherwise. Because they just two different columns of U. U^T*U is still the nxn identity matrix. You just can't call it the inverse. That's all I meant to clarify. You were doing fine otherwise.

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