Example of a symmetric matrix which doesn't have orthogonal eigenvectors

Question

I'm looking for an example of a symmetric matrix $A$ which doesn't have orthogonal eigenvectors.

Here's what I tried: I was able to prove that the eigenvectors corresponding to each distinct eigenvalue of a symmetric matrix are orthogonal. So, I realise that the example I'm looking for is a symmetric matrix with at least one repeated eigenvalue for which there are no orthogonal eigenvectors. But I'm not sure how to construct such an example.

Answer 1

So, I realise that the example I'm looking for is a symmetric matrix with at least one repeated eigenvalue for which there are no orthogonal eigenvectors.

If you ever get two of more linearly independent eigenvectors corresponding to the same eigenvalue, you can apply the Gram Schmidt process and end up with orthogonal vectors... which will continue being eigenvectors!

Thus, it is not possible to get any symmetric matrix which doesn't have orthogonal eigenvectors. (Note that I'm not saying that all eigenvectors will be orthogonal but that you can always find orthogonal ones.)

Answer 2

I proved that if A has distincts eigenvalues. Its eigenvectors are always orthogonal. Let $\alpha_1$ and $\alpha_2$ two distincts eigenvalues of $A$. So we have:

$(1)$ $$Ax_1=\alpha_1 x_1$$ $(2)$ $$Ax_2=\alpha_2 x_2$$ where $x_1$ and $x_2$ are eigenvectors.

Now we are multiplying $(1)$ with $x_2^T$, $(2)$ with $x_1^T$ and get: $$x_2^T A x_1=\alpha_1 x_2^T x_1 $$ $$x_1^T A x_2=\alpha_2 x_1^T x_2 $$

We transpoze both sides of first equation,and subtract from the second one and get: $$0=(\alpha_1 - \alpha_2)x_1^T x_2$$ As we chose eigenvalues are distincts, $x_1^T x_2$ must be zero. So finally we get: $x_1$ and $x_2$ orthogonal.

I thought that might help you to transmute your question to "Is there any symmetric matrix that has distinct eigenvectors for the same eigenvalue and are they orthogonal?"