# Orthogonal Eigenvectors

This seems a simple question but I can't find the solution by myself. Please help.

Say we have a 2 by 2 matrix with different eigenvalues. Corresponding to each eigenvalue, there are a number of eigenvectors.

Q1. Could the eigenvectors corresponding to the same eigenvalue have different directions?
Q2. Could the eigvenvectors corresponding to the same eigenvalue be orthogonal?

Q3. How can we prove that there is a pair of orthogonal eigenvalues for the 2 by 2 matrix, each for one eigenvalue?

Your help would be appreciated.

## Answers and Replies

This seems a simple question but I can't find the solution by myself. Please help.

Say we have a 2 by 2 matrix with different eigenvalues. Corresponding to each eigenvalue, there are a number of eigenvectors.

So there are 2 distinct eigenvalues. This means to me that there are 2 distinct eigenspaces each of dimension 1.

Q1. Could the eigenvectors corresponding to the same eigenvalue have different directions?
Q2. Could the eigvenvectors corresponding to the same eigenvalue be orthogonal?

No. The eigenspaces have dimension 1 in this case, so every two eigenvectors from the same eigenvalue are linearly dependent.

Q3. How can we prove that there is a pair of orthogonal eigenvalues for the 2 by 2 matrix, each for one eigenvalue?

We can't because it is not true. It is true for symmetric/hermitian matrices however.

Thanks a lot. I think its clear to me now. Bellow is my proof. I hope it's correct.

$Ax_{1}=\lambda_{1} x_{1}$
$Ax_{2}=\lambda_{2} x2$

multiplying the first equation with $x^{T}_{2}$ and the second one with $x^{T}_{1}$ , we have

$x^{T}_{2}Ax_{1}=\lambda_{1} x^{T}_{2} x_{1}$ (1)

$x^{T}_{1}Ax_{2}=\lambda_{2} x^{T}_{1} x_{2}$

transposoing the latter, we have

$x^{T}_{2}A^{T}x_{1}=\lambda_{2} x^{T}_{2} x_{1}$ (2)

if A is symmetric then (1) and (2) yields $x^{T}_{2} x_{1} =0$ or $\lambda_{1} = \lambda_{2}$

Do you have an intuitive reason as why for a symmetric matrix, two eigenvectors from distinct eigenspaces are orthogonal?