Question on operator's eigenvectors and eigenvalues?

Question

I'm reading a book on quantum mechanics titled "Quantum Mechanics: The Theoretical Minimum" by Leonard Susskind (if that's any help) I'm reading about quantum states and spin.. the author derives a 2x2 operator

σ_n = {(cosθ,sinθ),(sinθ,-cosθ)} this is supposed to be a matrix with entries

a11: cosθ
a12: sinθ
a21: sinθ
a22: -cosθ

and he asks the reader to find its eigenvalues and corresponding eigenvectors.  I've manages to find the eigenvalues (λ=+1 and λ=-1) obvious result for spin. The author hints the reader to use an arbitrary column eigenvector 
|A> = (cosα,sinα)
and solve for α in terms of θ to find the eigenvectors.  Somehow, the answer is the book is

λ=+1
( cos(½θ) , sin(½θ) )

and

λ=-1 
( -sin(½θ) , cos(½θ) )

I don't know how he arrives to these results.  Even wolfram alpha gives different eigenvectors for eigenvalues +1 and -1:

λ=+1
(cotθ - 1 , 1)

and

λ=-1
(cotθ + 1 , 1)

Randy P · Answer

Two comments.
1. When you have an eigenvector, you can multiply it by any complex constant and get another eigenvector for the same eigenvalue. Eigenvectors aren't unique. So one thing to explore is whether you can get the textbook's vector from Wolfram's vector by such a multiplication.

2. The book wasn't suggesting you solve for an arbitrary eigenvector, it was specifically saying "use (cos alpha, sin alpha) as a trial eigenvector and solve for alpha".

When you do that, you get
cos(theta) cos(alpha) + sin(theta) sin(alpha) = +-cos(alpha)
sin(theta) cos(alpha) - cos(theta) sin(alpha) = +-sin(alpha)

Not that I know how to proceed without some algebra and some thought, but I assume that this will lead to something like theta = 2*alpha by way of an equation like sin(theta) = 2 sin(alpha) cos(alpha).

In fact you might try to solve for sin(theta) by elimination and see if that's what you get.

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Question on operator's eigenvectors and eigenvalues?

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