Question

Diagonalize a if possible. (find p and d such that a = pdp−1 for the given matrix

Answer 1

$\mathbf A=\begin{bmatrix}-10&30\\-6&17\end{bmatrix}$

Compute the eigenvalues of $\mathbf A$:

$\begin{vmatrix}-10-\lambda&30\\-6&17-\lambda\end{vmatrix}=(-10-\lambda)(17-\lambda)+180=(\lambda-5)(\lambda-2)=0$

$\implies\lambda=5,\lambda=2$

Find the corresponding eigenvectors $\eta$:

$\lambda_1=2\implies\begin{bmatrix}-12&30\\-6&15\end{bmatrix}\eta_1=\mathbf0$

$\implies\eta_1=\begin{bmatrix}5\\2\end{bmatrix}$

$\lambda_2=5\implies\begin{bmatrix}-15&30\\-6&12\end{bmatrix}\eta_2=\mathbf0$

$\implies\eta_2=\begin{bmatrix}2\\1\end{bmatrix}$

Now,

$\mathbf A=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}\mathrm{diag}(\lambda_1,\lambda_2)\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^{-1}$

$\mathbf A=\begin{bmatrix}5&2\\2&1\end{bmatrix}\begin{bmatrix}2&0\\0&5\end{bmatrix}\begin{bmatrix}5&2\\2&1\end{bmatrix}^{-1}$