# Orthogonality relation of eigenvectors for a self-adjoint operator

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.

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