Why eigenvector of Symmetric Matrix to be orhogonal for repeated eigenvalues.

Question

I know that for a symmetric matrix, its eigenvector are orthogonal. But for a repeated eigenvalue for a symmetric matrix, why still its eigenvalue must be still orthogonal.

I read somewhere that symmetric matrix can not have defective eigenvalue, so for a repeated eigenvalue with m multiplicity, we will still have m orthogonal eigenvector for it.

Kindly please for your answer, can you show by taking a 3 by 3 matrix and show it.

Answer 1

If a symmetric matrix $A$ has an eigenvalue $\lambda$ with multiplicity $k$, then the dimension of $\ker(A-\lambda I)$ is equal to $k$. Therefore, you can find an orthogonal basis for $\ker(A-\lambda I)$ (because all subspaces have an orthogonal basis).

Now note that

1. The dimension of the kernel is $k$, therefore the basis has $k$ elements.
2. All elements of $\ker(A-\lambda I)$ are eigenvectors of $A$.

From the points above, you can conclude that the orthogonal basis of $\ker(A-\lambda I)$ is comprised of $k$ orthogonal eigenvectors of $A$.