Example4.3.8

Let's consider the matrix \(A = \left[\begin{array}{rr} -2 \amp 2 \\ -5 \amp 4 \\ \end{array}\right]\) whose characteristic equation is

\begin{equation*} \det(A-\lambda I) = (-2-\lambda)(4-\lambda)+10 = 2 - 2\lambda + \lambda^2 = 0 \text{.} \end{equation*}

Applying the quadratic formula to find the eigenvalues, we obtain

\begin{equation*} \lambda = \frac{2\pm\sqrt{(-2)^2-4\cdot1\cdot2}}{2}=1\pm i \text{.} \end{equation*}

Here we see that the matrix \(A\) has two complex eigenvalues and is therefore not diagonalizable.

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