Solution EE.C27 Chris Black

Since we are given that the characteristic polynomial of $A$ is $\charpoly{A}{x} = (x+2)(x-2)^2(x-4)$, we see that the eigenvalues are $\lambda = -2$, $\lambda = 2$ and $\lambda = 4$. The eigenspaces are \begin{align*} \eigenspace{A}{-2}&= \spn{\colvector{0\\0\\1\\1}}\\ \eigenspace{A}{2} &= \spn{\colvector{1\\1\\2\\0},\colvector{3\\1\\0\\2}}\\ \eigenspace{A}{4} &= \spn{\colvector{1\\1\\1\\1}} \end{align*}