Proving the Dot Product of Orthonormal Vectors in a Matrix

[Lonely Lemon]

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...

 

[Inferior89]

Do you know anything about inverses of orthonormal matrices?

 

[Lonely Lemon]

I know that if U is orthogonal then (U^T)(U)=I, and by the Invertible Matrix Theorgem U inverse = U transpose?

 

[Inferior89]

Yes, use this together with Ux = x^T U^T, matrix multiplication is associative and that x . u = x^T u.

 

[Lonely Lemon]

Okay, I get this far:
(Ux).(Uy)=(x^T)(U^T).(y^T)(U^T)

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

 

[Inferior89]

You did one steep too far.

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

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

 

[Dick]

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.

 

[Lonely Lemon]

Oh, then we have (x^T)(y)=x.y

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

 

[Inferior89]

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.

 

[Dick]

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.