##### 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.

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