Find an invertible matrix \(Q\) so that \(Q^{-1} \begin{pmatrix} -1 & 1 \\ 0 & 2 \end{pmatrix} Q\) is diagonal.
Solution
1. Let\[ A=\begin{pmatrix}-1&1\\0&2\end{pmatrix}. \]Its eigenvalues are \(-1\) and \(2\).
Because \(A\) is upper triangular, its eigenvalues are its diagonal entries.
2. An eigenvector for \(\lambda=-1\) is\[ v_{-1}=\begin{pmatrix}1\\0\end{pmatrix}. \]
Directly, \[ Av_{-1}=\begin{pmatrix}-1\\0\end{pmatrix}=-v_{-1}. \]
3. An eigenvector for \(\lambda=2\) is\[ v_2=\begin{pmatrix}1\\3\end{pmatrix}. \]
Solving \((A-2I)v=0\) gives \[ \begin{pmatrix}-3&1\\0&0\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}=0, \] so \(y=3x\); taking \(x=1\) gives \(v_2=(1,3)^t\).
4. Set\[ Q=\begin{pmatrix}1&1\\0&3\end{pmatrix}. \]Then \(Q\) is invertible and\[ \boxed{Q^{-1}AQ=\begin{pmatrix}-1&0\\0&2\end{pmatrix}.} \]
The columns of \(Q\) are the eigenvectors from 2 and 3, and \(\det Q=3\ne0\). Hence \[ AQ=Q\begin{pmatrix}-1&0\\0&2\end{pmatrix}, \] and multiplying on the left by \(Q^{-1}\) gives the displayed diagonalization.