Proofs from Diagonalisation

Let $v_1,\ldots,v_n$ be eigenvectors for distinct eigenvalues $\lambda_1,\ldots,\lambda_n$ and suppose

To isolate one coefficient, fix $k$ and apply

to both sides. The factors commute because each is a polynomial in $A$. Every term except the $k$th is killed by one factor, while

Thus the transformed relation is

The scalar product is non-zero and $v_k\ne0$, so $c_k=0$. Since $k$ was arbitrary, all coefficients vanish.

If $v_1,\ldots,v_n$ are independent eigenvectors, let

The matrix $P$ is invertible and applying $A$ to each column gives $AP=PD$. Hence $P^{-1}AP=D$.

Conversely, if $P^{-1}AP=D$, then $AP=PD$. The columns of $P$ are therefore eigenvectors corresponding to the diagonal entries of $D$. Since $P$ is invertible, its columns are linearly independent.