###### Example 4.3.5

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.