Let $M$ be a $3 \times 3$ matrix with eigenvalues $\lambda_1 = -1.2, \ \lambda_2 = 0.5 \ \lambda_3 = 2$ with corresponding eigenvectors $$\mathbf{v}_1 = \left[\begin{array}{c} 2\cr -2\cr 0 \end{array}\right], \mathbf{v}_2= \left[\begin{array}{c} 0\cr -1\cr -1 \end{array}\right], \mathbf{v}_3= \left[\begin{array}{c} 0\cr 1\cr 2 \end{array}\right].$$ Consider the difference equation

$$\mathbf{x}_{k+1} = M\mathbf{x}_k$$ with initial condition $\mathbf{x}_0 = \left[\begin{array}{c} 3\cr 3\cr 3 \end{array}\right]$.

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Write the initial condition as a linear combination of the eigenvectors of $M$.

That is, write $\mathbf{x}_0 = c_1 \mathbf{v_1} + c_2 \mathbf{v_2} + c_3 \mathbf{v_3}$ $=$ $\mathbf{v_1} +$$\mathbf{v_2} +$$\mathbf{v_3}$

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In general, $\mathbf{x}_k =$ $\big($ $\big)^k \ \mathbf{v}_1+$ $\big($ $\big)^k \ \mathbf{v}_2+$ $\big($ $\big)^k \ \mathbf{v}_3$

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Specifically, $\mathbf{x}_{4} =$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$

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For large $k$, $\mathbf{x}_{k} \approx$ $\big($ $\big)^k\ $ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$