## Proof That Distinct Eigenvalues Return Distinct Eigenvectors

Theorem
If
$\lambda_1 , \lambda_2$
are eigenvalues of a square matrix
$A$
with
$\lambda_ 1 \neq \lambda_2$
and
$\mathbf{v}_1 , \: \mathbf{v}_2$
are the eigenvectors associated with
$\lambda_1 , \: \lambda_2$
respectively, then
$\mathbf{v}_1 \neq \mathbf{v}_2$
.
Proof
$\mathbf{v}_1 , \lambda_1$
satisfy
$A \mathbf{v}_1 = \lambda_1 \mathbf{v}_1$

$\mathbf{v}_2 , \lambda_2$
satisfy
$A \mathbf{v}_2 = \lambda_2 \mathbf{v}_2$

Set
$\mathbf{v}_2 = \mathbf{v}_1$
and subtract the first from the second.
$A( \mathbf{v}_1 - \mathbf{v}_1) = (\lambda_2 -\lambda_1 ) \mathbf{v}_1$

The left hand side is zero. The right hand side is then also zero, and since
$\mathbf{v}_1 \neq 0$
we must have
$\lambda_1 = \lambda_2$
. This is a contradiction, so distinct eigenvalues give rise to distinct eigenvectors.
The converse is not true. Distinct eigenvectors can have the same eigenvalues.
The matrix
$\left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right|$
has eigenvectors
$\begin{pmatrix}1\\0\end{pmatrix} , \: \begin{pmatrix}0\\1\end{pmatrix}$
corresponding to the eigenvalue
$\lambda =1$
.