Prove l*conj(l)=1 for Orthogonal Matrix A

[phrygian]

Homework Statement

Let l be an eigenvalue of an orthogonal matrix A, where l = r + is. Prove that l * conj(l) = r^2 + s^2 = 1.

Homework Equations

The Attempt at a Solution

I am really confused on where to go with this one.

I have Ax = A I x = A A^T A x = l^3 x

and Ax = l x so l x = l^3 x

l = l^3

l^2 = 1
l = 1 or -1

But I can't really figure out what to do from here, am I even on the right track?

Thanks for the help

 

[fanxiu]

Hi, i am just studying linear algebra (the final is coming next week, so stressed!).

I think the question about l*conjugate(l) = r^2+s^2 = 1 just mean the length of l is 1

But i think if Matrix with complex entries, say M, is orthogonal means M is unitary. So the length of every eigenvalue is 1

since unitary matrix won't change vector's langth so its eigenvalues' length is always 1

 

[D H]

But i think if Matrix with complex entries, say M, is orthogonal means M is unitary. So the length of every eigenvalue is 1

since unitary matrix won't change vector's langth so its eigenvalues' length is always 1

phrygian cannot use this fact; he/she has to prove it.

phrygian, the cube route isn't really going to help here. You went one step too far.

You know that $$\mathbf A \vec x = \lambda \vec x$$. The conjugate transpose of the right-hand side is $$(\lambda \vec x)^* = \vec x^*\lambda^*$$. What is the matrix product of this conjugate transpose with $$\lambda \vec x$$?

 

[phrygian]

Is there a way to do this without using conjugate transposes? The book has a hint that says first show that ||Ax|| = ||x|| for any vector x, but i just can't seem to get past that first step

 

[HallsofIvy]

What is the precise definition of "orthogonal" matrix? Some texts use that a matrix, Q, is orthogonal if and only if <Qu, v>= <u, Qv> for all vector u and v (< u, v> is the inner product). From that it is not too hard to show that ||Qv||= ||v||.