Let \(A = \begin{pmatrix} 7 & -3 \\ 1 & 3 \end{pmatrix}\). Find an invertible matrix \(P\) and a diagonal matrix \(D\) with \(P^{-1}AP = D\). [You should not have to compute \(P^{-1}\).]
Solution
The characteristic polynomial is \[ \det(\lambda I-A)=(\lambda-7)(\lambda-3)+3 =\lambda^2-10\lambda+24=(\lambda-4)(\lambda-6). \] For \(\lambda=4\), an eigenvector is \((1,1)^t\); for \(\lambda=6\), an eigenvector is \((3,1)^t\). Thus \[ P=\begin{pmatrix}1&3\\1&1\end{pmatrix},\qquad D=\begin{pmatrix}4&0\\0&6\end{pmatrix}. \] Since \(\det P=-2\ne0\), \(P\) is invertible, and because its columns are eigenvectors in the displayed order, \[ P^{-1}AP=D. \]