 Solution

An $n\times n$ matrix is diagonalizable over a chosen field exactly when it has a basis of $n$ linearly independent eigenvectors. Equivalently, $P^{-1}AP$ is diagonal for some invertible $P$. Distinct eigenvalues suffice but are not necessary. For example,is not diagonalizable: its only eigenvalue is $1$ and its eigenspace is the one-dimensional span of $(1,0{)}^T$. It is a nontrivial Jordan block.

A Hermitian matrix satisfies $A^{†}=A$, where the dagger denotes conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenThe denominator is positive real, soFor $Av=\lambda v$ and $Aw=\mu w$, Hermiticity also givesThusThis is the necessary qualification to the PDF's orthogonality wording: arbitrary eigenvectors belonging to a repeated eigenvalue need not be orthogonal. Within each repeated eigenspace, Gram-Schmidt orthogonalization produces an orthonormal basis; together the eigenspaces give an orthonormal eigenbasis by the finite-dimensional spectral theorem. For example, $(1,0{)}^T$ and $(1,1{)}^T$ are nonorthogonal eigenvectors of the identity matrix.