= 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,
$$
J=\begin{pmatrix}1&1\\0&1\end{pmatrix}
$$
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 $A^\dagger=A$, where the dagger denotes <conjugate transpose>. If $Av=\lambda v$ with $v\ne0$, then
$$
\lambda=\frac{v^\dagger Av}{v^\dagger v},\qquad
(v^\dagger Av)^*=v^\dagger A^\dagger v=v^\dagger Av.
$$
The denominator is positive real, so
$$
\boxed{\lambda\in\mathbb R.}
$$
For $Av=\lambda v$ and $Aw=\mu w$, Hermiticity also gives
$$
\mu v^\dagger w=v^\dagger Aw=(Av)^\dagger w=\lambda v^\dagger w.
$$
Thus
$$
\boxed{\lambda\ne\mu\quad\Longrightarrow\quad v^\dagger w=0.}
$$
This 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>.
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