# Recitation 8B

Problem 1: Is \mathbf{x}=\begin{bmatrix}3\\-2\\1\end{bmatrix} an eigenvector of A=\begin{bmatrix}-4&3&3\\2&-3&-2\\-1&0&-2\end{bmatrix}?

Solution: Because A\mathbf{x}=\begin{bmatrix}-15\\10\\-5\end{bmatrix}=(-5)\mathbf{x}, it is an eigenvector.

Problem 2: Is \lambda=4 an eigenvalue of A=\begin{bmatrix}3&0&-1\\2&3&1\\-3&4&5\end{bmatrix}.

Solution: Because A-\lambda I=\begin{bmatrix}-1&0&-1\\2&-1&1\\-3&4&1\end{bmatrix} is not invertible, it is an eigenvalue.

Problem 3: Find a basis for the eigenspace corresponding to \lambda=-5 of A=\begin{bmatrix}-4&1&1\\2&-3&2\\3&3&-2\end{bmatrix}.

Solution: It is enough to find a basis for the solution set of (A-\lambda I)\mathbf{x}=\mathbf{0}. By row reduction, we obtain \begin{bmatrix}1&1&1\\2&2&2\\3&3&3\end{bmatrix}\sim \begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}. Therefore the parametric vector form of the solutions is \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}-1\\1\\0\end{bmatrix}x_2+\begin{bmatrix}-1\\0\\1\end{bmatrix}x_3. In other words, a basis can be \begin{bmatrix}-1\\1\\0\end{bmatrix}, \begin{bmatrix}-1\\0\\1\end{bmatrix}.

Problem 4: If A^2 is the zero matrix, then the only eigenvalue of A is 0.

Solution: Suppose \lambda is an eigenvalue of A and \mathbf{x}\neq \mathbf{0} that satisfies A\mathbf{x}=\lambda\mathbf{x}. Now we have \mathbf{0}=A^2\mathbf{x}=A(A\mathbf{x})=A(\lambda\mathbf{x})=\lambda(A\mathbf{x})=\lambda(\lambda\mathbf{x})=\lambda^2\mathbf{x}. Since \mathbf{x}\neq 0, \lambda^2=0, which implies \lambda=0.

Problem 5: Find the characteristic equation of \begin{bmatrix}4&0&-1\\0&4&-1\\1&0&2\end{bmatrix}.

Solution: The characteristic polynomial is \mathrm{det}\left(\lambda I - A\right), which is equal to \begin{aligned}\begin{vmatrix}\lambda-4 & 0 & 1 \\ 0 & \lambda-4 & 1 \\ -1 & 0 & \lambda-2\end{vmatrix} &=(\lambda-4)\begin{vmatrix}\lambda-4 & 1 \\ 0 & \lambda - 2\end{vmatrix}+\begin{vmatrix}0 & \lambda-4 \\ -1 & 0\end{vmatrix}\\&=(\lambda-4)(\lambda-4)(\lambda-2)+(\lambda-4)\\&=(\lambda-4)(\lambda-3)^2.\end{aligned} Therefore, the characteristic equation is (\lambda-4)(\lambda-3)^2=0.