Let \(T\) be the linear operator on \(\RR^3\) which is represented in the standard ordered basis by the matrix \[ \begin{bmatrix} -9 & 4 & 4 \\ -8 & 3 & 4 \\ -16 & 8 & 7 \end{bmatrix}. \] Prove that \(T\) is diagonalizable by exhibiting a basis for \(\RR^3\), each vector of which is a characteristic vector of \(T\).
Solution
Let \[ A=\begin{bmatrix}-9&4&4\\-8&3&4\\-16&8&7\end{bmatrix}. \] Its characteristic polynomial is \[ \chi_A(x)=(x-3)(x+1)^2. \] For \(\lambda=-1\), direct multiplication shows \[ A(1,2,0)^t=-(1,2,0)^t, \qquad A(1,0,2)^t=-(1,0,2)^t. \] For \(\lambda=3\), \[ A(1,1,2)^t=3(1,1,2)^t. \] The three vectors \[ v_1=(1,2,0),\qquad v_2=(1,0,2),\qquad v_3=(1,1,2) \] are linearly independent, since \[ \det\begin{bmatrix}1&1&1\\2&0&1\\0&2&2\end{bmatrix}=-2\ne0. \] Therefore they form a basis of \(\mathbb R^3\) consisting of characteristic vectors, so \(T\) is diagonalizable.