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 \[\lambda =1\]
.