##### Example4.3.5

We will try to find a diagonalization of \(A = \left[\begin{array}{rr} 0 \amp 4 \\ -1 \amp 4 \\ \end{array}\right] \text{.}\)

Once again, we find the eigenvalues by solving the characteristic equation:

\begin{equation*} \det(A-\lambda I) = -\lambda(4-\lambda) + 4 = (2-\lambda)^2 = 0 \text{.} \end{equation*}In this case, there is a single eigenvalue \(\lambda=2\text{.}\)

We find a basis for the eigenspace \(E_2\) by describing \(\nul(A-2I)\text{:}\)

\begin{equation*} A-2I = \left[\begin{array}{rr} -2 \amp 4 \\ -1 \amp 2 \\ \end{array}\right] \sim \left[\begin{array}{rr} 1 \amp -2 \\ 0 \amp 0 \\ \end{array}\right] \text{.} \end{equation*}This shows that the eigenspace \(E_2\) is one-dimensional with \(\vvec_1=\twovec{2}{1}\) forming a basis.

In this case, there is not a basis of \(\real^2\) consisting of eigenvectors of \(A\text{,}\) which tells us that \(A\) is not diagonalizable.