# Eigenspaces-symmetric matrix (2x2)

1. Dec 19, 2006

### *best&sweetest*

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!

2. Dec 20, 2006

### matt grime

You know a lot more than this. You know there are two eigenvectors and two eigenvalues. Now, what is the only way we have of deciding when two vectors are orthogonal? Remember, symmetrice means A=A^t, and what have you been taught about matrices and the only method you know for deciding when two vectors are orthogonal.

Last edited: Dec 20, 2006