Question

Prove: Every eigenvalue of an orthogonal transformation has absolute value 1.
Please, help. I read the answer from https://answers.yahoo.com/question/index?qid=20110210040434AAJMMtz
But don't understand. Explain me, please

Answer 1

@Zarkon

Answer 2

What part of it is confusing? When he wrote (At) I think what he meant was A transposed.

Answer 3

If Ax = λx then \((A^t)Ax = (At)λx: that is \((\dfrac{1}{λ})x = A^tx\), so 1/λ is an eigenvalue of \(A^t\)

Answer 4

why \(\dfrac{1}{\lambda}x= A^tx\)?

Answer 5

He actually explains that in the second paragraph of the answer.

Answer 6

he said A and \(A^t\) has the same eigen value \(\lambda\), where 1/lambda come from?

Answer 7

from the 1st paragraph

Answer 8

(At)Ax = (At)λx: that is (1/λ)x = At*x

Is just algebra since (At)A=I

Answer 9

\[A^tA=I\]

so \[Ax=\lambda x\]

mult by \(A^t\)

\[A^tAx=A^t(\lambda x)\]
\[x=\lambda A^tx\]

\[\frac{1}{\lambda}x=A^tx\]

Answer 10

Thank you so much. Sorry for bothering by the dummy question. :)

Answer 11

All good. =)

Answer 12

Thank you, see the comment!! it sounds like the proof is not done yet, right?

Answer 13

I think the comment is wrong, since the second paragraph of the answer explains why A and At have the same eigenvalues.

Answer 14

Thanks again