Orthogonality relation of eigenvectors for a self-adjoint operator

Question

So everyone knows eigenvectors corresponding to different eigenvalues are orthogonal to each other, given that the operator is self-adjoint.

If we have a self-adjoint operator, say $L$, is it possible that $\exists u, v$ such that $Lu=\lambda u$, $Lv=\lambda v$ and $\langle u, v\rangle=0$. In other words, we have eigenvectors with the same eigenvalues to $L$ and they are still orthogonal?

This is motivated by considering the angular momentum operators in Quantum Mechanics, I was wondering if there is a simpler example in Linear Algebra.

Answer 1

Just take the $L$ to be the identity operator on a finite dimensional Hilbert space (inner product space). And use the orthonormal basis basis .

If your eigen-space has dimension greater than $1$ then it is always possible. Just take the orthonormal basis of the eigen-space.

Answer 2

Yes, it is possible.

In fact, if an operator $L$ (not necessarily self-adjoint) has two linearly independent eigenvectors associated with $\lambda$ (that is, the eigenspace associated with $\lambda$ has dimension at least $2$), then we can always find a pair of orthonormal eigenvectors associated with $\lambda$ by applying the Gram-Schmidt process to $\{u,v\}$.

Some examples: consider $$ L = \pmatrix{1&0\\0&1}, \quad L = \pmatrix{1&1&1\\1&1&1\\1&1&1}, \quad L= \pmatrix{1&0&-1\\0&1&-2\\ 0&0&-3}. $$