Diagonalize the matrix \(A = \begin{bmatrix} 1 & 2 \\ 4 & -1 \end{bmatrix}\) and use the diagonal form of \(A\) to compute \(A^2\).
Solution
1. The characteristic polynomial of \(A\) is\[ \chi_A(\lambda)=\det(\lambda I-A)=\lambda^2-9. \]
Directly, \[ \det\begin{bmatrix}\lambda-1&-2\\-4&\lambda+1\end{bmatrix} =(\lambda-1)(\lambda+1)-8 =\lambda^2-9. \]
2. The eigenvalues are \(3\) and \(-3\), with eigenvectors\[ v_+=\begin{bmatrix}1\\1\end{bmatrix}, \qquad v_-=\begin{bmatrix}1\\-2\end{bmatrix}. \]
For \(\lambda=3\), the equation \((A-3I)v=0\) gives \(-2x+2y=0\), hence \(y=x\). For \(\lambda=-3\), the equation \((A+3I)v=0\) gives \(4x+2y=0\), hence \(y=-2x\).
3. Therefore\[ P=\begin{bmatrix}1&1\\1&-2\end{bmatrix}, \qquad D=\begin{bmatrix}3&0\\0&-3\end{bmatrix} \]satisfy\[ A=PDP^{-1}. \]
2. Since they correspond to distinct eigenvalues, they are linearly independent, so \(P\) is invertible. The relation \(AP=PD\) is exactly the eigenvector equations, and hence \(A=PDP^{-1}\).
4. Using the diagonal form,\[ A^2=PD^2P^{-1}=P(9I_2)P^{-1}=9I_2 =\begin{bmatrix}9&0\\0&9\end{bmatrix}. \]