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*best&sweetest*
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I have a tough time proving that if a 2x2 symmetric matrix with real entries has two distinct eigenvalues than the eigenspaces corresponding to those eigenvalues are perpendicular lines through the origin in R^2.
All I have is symmetric matrix
a b
b d
and I know that since it has distinct eigenvalues it must be that the discriminant of the characteristic polynomial must be greater than zero, i.e
(a-d)^2 + 4b^2 > 0
so it cannot be that a=d and b=0 at the same time.
Now, I have no clue what to do next, and I have a feeling that this might be on my final exam, so any help would be greatly appreciated. Thanks!
All I have is symmetric matrix
a b
b d
and I know that since it has distinct eigenvalues it must be that the discriminant of the characteristic polynomial must be greater than zero, i.e
(a-d)^2 + 4b^2 > 0
so it cannot be that a=d and b=0 at the same time.
Now, I have no clue what to do next, and I have a feeling that this might be on my final exam, so any help would be greatly appreciated. Thanks!