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Homework Statement
Show that the matrix
A = [cos θ -sin θ
sin θ cos θ]
will have complex eigenvalues if θ is not a multiple of π. Give a geometric interpretation of this result.
Homework Equations
Ax = λx, so
det(A-λI) = 0
The Attempt at a Solution
In this case,
A-λI = [(cos θ - λ) -sin θ
sin θ (cos θ - λ)]
so calculating the determinate, I have:
[itex]det(A-λI) = (cos θ - λ)(cos θ - λ) + sin^{2} θ [/itex]
[itex]= cos^{2} θ - 2λcos θ + λ^{2} + sin^{2} θ[/itex]
[itex]= λ^{2} - 2λcos θ + 1[/itex]
Setting this polynomial to zero, I have
[itex]
p(λ) = λ^{2} - 2λcos θ + 1 = 0
[/itex]
When cos θ is not a multiple of π, I need to use quadratic formula to solve this equation. Doing so, I get:
[itex]
λ = cos θ \pm sin θ
[/itex]
However, this value is not complex, is it? I'm a bit confused on what's going on.