Multiplying both sides of \(P^{-1}AP = D\) on the left by \(P\) gives \(AP = PD\text{.}\) Expanding column by column:
\begin{align*}
A[v_1 \, v_2 \, \cdots \, v_n]\amp = [v_1 \, v_2 \, \cdots \,v_n]\begin{bmatrix}
\lambda_1 \amp 0 \amp \cdots \amp 0 \\
0 \amp \lambda_2 \amp \cdots \amp 0 \\
\vdots \amp \vdots \amp \ddots \amp \vdots \\
0 \amp 0 \amp \cdots \amp \lambda_n
\end{bmatrix}\\
[Av_1 \, Av_2 \, \cdots \, Av_n]\amp = [\lambda_1 v_1 \, \lambda_2 v_2 \, \cdots \, \lambda_n v_n]
\end{align*}
Comparing columns gives \(Av_i = \lambda_i v_i\) for \(i = 1, 2, \ldots, n\text{.}\) So each column \(v_i\) of \(P\) is an eigenvector of \(A\) with eigenvalue \(\lambda_i\text{.}\) Since \(P\) is invertible, its columns are linearly independent. Therefore \(A\) has \(n\) linearly independent eigenvectors.
