Find the eigenvalues and eigenvectors of the matrix \(A = \begin{bmatrix} 1 & -2 \\ 3 & -4 \end{bmatrix}\).
Solution
The characteristic polynomial is \[ \det(A-\lambda I) =\det\begin{bmatrix}1-\lambda&-2\\3&-4-\lambda\end{bmatrix} =(\lambda+1)(\lambda+2). \] Thus the eigenvalues are \(-1\) and \(-2\).
For \(\lambda=-1\), \[ A+I=\begin{bmatrix}2&-2\\3&-3\end{bmatrix}, \] so the eigenspace is \[ \ker(A+I)=\operatorname{span}\{(1,1)^T\}. \] For \(\lambda=-2\), \[ A+2I=\begin{bmatrix}3&-2\\3&-2\end{bmatrix}, \] so \[ \ker(A+2I)=\operatorname{span}\{(2,3)^T\}. \] Hence every nonzero multiple of \((1,1)^T\) is an eigenvector for \(-1\), and every nonzero multiple of \((2,3)^T\) is an eigenvector for \(-2\).