= Solution
Take <eigenvectors> $v_i,v_j$ for distinct <eigenvalues> $\lambda_i,\lambda_j$. Since the <eigenvalues> are real and $A$ is a <Hermitian matrix>,
$$
\lambda_jv_i^\dagger v_j=v_i^\dagger Av_j=(Av_i)^\dagger v_j=\lambda_i v_i^\dagger v_j.
$$
Hence $(\lambda_j-\lambda_i)v_i^\dagger v_j=0$, so \b[<eigenvectors> for distinct <eigenvalues> are <orthogonal vectors>]. They are therefore <linearly independent>; after normalization the given $n$ <eigenvectors> form an <orthonormal basis> of $\mathbb C^n$.
Back to article page
