Wikipedia states:
Moreover, a Hermitian matrix has orthogonal eigenvectors for distinct eigenvalues. Even if there are degenerate eigenvalues, it is always possible to find an orthogonal basis of $\mathbb C_n$ consisting of $n$ eigenvectors of $A$.
What confuses me about this statement is that it says "it is always possible ...". My understanding is that either a matrix has an orthogonal basis of eigenvectors, or it doesn't. If it does, then we can only find an orthogonal basis of eigenvectors (i.e. we cannot find a non-orthogonal one), and if it doesn't, we can't.
The reason I think this, is because if we have eigenvectors $v$ and $w$, then if $v$ and $w$ are not orthogonal, there exists no scalar $a$ such that $v$ and $a\cdot w$ are suddenly orthogonal.
Where am I wrong in my understanding?
ps. I have very little experience with complex-valued mmatrices, as I've just started studying them.
Consider the identity matrix. Then any basis of $\mathbb{C}_n$ is a basis of eigenvectors, and it need not be orthogonal.
More generally, the wiggle room is that if an eigenvalue has multiplicity more than one, then you are free to choose any basis within that eigenvalue's eigenspace, and it need not be orthogonal. In your example, you may have $v$ and $w$ not orthogonal, but it may be that $v$ and $v+aw$ are orthogonal.