Find an invertible matrix \(A\) and a diagonal matrix \(B\) such that \(\begin{pmatrix} 8 & 9 \\ -6 & -7 \end{pmatrix} = ABA^{-1}\).
Solution
1. Let \(M = \begin{pmatrix} 8 & 9 \\ -6 & -7 \end{pmatrix}\); its eigenvalues are \(-1\) and \(2\).
\(\det(M - \lambda I) = (8-\lambda)(-7-\lambda) + 54 = \lambda^2 - \lambda - 2 = (\lambda + 1)(\lambda - 2)\).
2. An eigenvector for \(\lambda = -1\) is \(\begin{pmatrix} -1 \\ 1 \end{pmatrix}\).
\(M\begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -8 + 9 \\ 6 - 7 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} = -\begin{pmatrix} -1 \\ 1 \end{pmatrix}\).
3. An eigenvector for \(\lambda = 2\) is \(\begin{pmatrix} -3 \\ 2 \end{pmatrix}\).
\(M\begin{pmatrix} -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -24 + 18 \\ 18 - 14 \end{pmatrix} = \begin{pmatrix} -6 \\ 4 \end{pmatrix} = 2\begin{pmatrix} -3 \\ 2 \end{pmatrix}\).
4. Let \(A = \begin{pmatrix} -1 & -3 \\ 1 & 2 \end{pmatrix}\) (columns are the eigenvectors) and \(B = \begin{pmatrix} -1 & 0 \\ 0 & 2 \end{pmatrix}\).
definition.
5. Then \(M = A B A^{-1}\).
\(A\) has the eigenvectors as columns, so \(A^{-1} M A = B\), equivalently \(M = A B A^{-1}\).
6. Q.E.D.
5.