Question

Let \(A = \left( {\begin{aligned}{*{20}{c}}{.4}&{ - .3}\\{.4}&{1.2}\end{aligned}} \right)\). Explain why \({A^k}\) approaches \(\left( {\begin{aligned}{*{20}{c}}{ - .5}&{ - .75}\\1&{1.5}\end{aligned}} \right)\) as \(k \to \infty \).

Short answer

This implies that \(\mathop {\lim }\limits_{k \to \infty } {A^k} \to \left( {\begin{aligned}{*{20}{c}}{ - .5}&{ - .75}\\1&{1.5}\end{aligned}} \right)\).

Step-by-step solution

Step 1: Find the characteristic polynomial

Consider the matrices\(A = \left( {\begin{aligned}{*{20}{c}}{.4}&{ - .3}\\{.4}&{1.2}\end{aligned}} \right)\).

\(\begin{aligned}{c}\det \left( {A - \lambda I} \right) &= \det \left( {\begin{aligned}{*{20}{c}}{.4 - \lambda }&{ - .3}\\{.4}&{1.2 - \lambda }\end{aligned}} \right)\\ &= \left( {.4 - \lambda } \right)\left( {1.2 - \lambda } \right) + 1.2\\ &= {\lambda ^2} - 1.6\lambda + .6\\ &= \left( {\lambda - 1} \right)\left( {\lambda - .6} \right)\end{aligned}\)

Thus, the eigenvalues are \(1\) and \(0.6\).

Find the Eigenvector for \(\lambda  = {\bf{1}}\)

\(\begin{aligned}{c}\left( {A - \lambda I} \right) &= \left( {\begin{aligned}{*{20}{c}}{ - .6}&{ - .3}\\{.4}&{.2}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}2&1\\2&1\end{aligned}} \right)\;\left\{ {{R_1} \to \frac{1}{{ - 3}}{R_1},\;{R_2} \to \frac{1}{2}{R_2}} \right\}\\ &= \left( {\begin{aligned}{*{20}{c}}2&1\\0&0\end{aligned}} \right)\;\left\{ {{R_1} \to {R_2} - {R_1}} \right\}\end{aligned}\)

The general solution is as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = {x_1}\left( {\begin{aligned}{*{20}{c}}1\\{ - 2}\end{aligned}} \right)\)

Therefore, the eigenvector for \(\lambda = 1\) is \({v_1} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\).

Find the Eigenvector for \(\lambda  = {\bf{0}}{\bf{.6}}\)

\(\begin{aligned}{c}\left( {A - .6I} \right) &= \left( {\begin{aligned}{*{20}{c}}{ - .2}&{ - .3}\\{.4}&{.6}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{ - .2}&{ - .3}\\{.2}&{.3}\end{aligned}} \right)\;\left\{ {{R_2} \to \frac{1}{2}{R_2}} \right\}\\ &= \left( {\begin{aligned}{*{20}{c}}2&3\\0&0\end{aligned}} \right)\;\left\{ {{R_2} \to {R_2} + {R_1},{R_1} \to \frac{1}{{ - .1}}{R_1}} \right\}\end{aligned}\)

The general solution is as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = {x_1}\left( {\begin{aligned}{*{20}{c}}1\\{ - \frac{2}{3}}\end{aligned}} \right)\)

Therefore, the eigenvector for \(\lambda = - 3\) is \({v_2} = \left( {\begin{aligned}{*{20}{c}}{ - 3}\\2\end{aligned}} \right)\).

Find the matrix \(P\)

The matrix\(P\)is formed by eigenvectors,

\(P = \left( {\begin{aligned}{*{20}{c}}{ - 1}&{ - 3}\\2&2\end{aligned}} \right)\)

Now find the inverse of the matrix\(P\).

\(\begin{aligned}{c}{P^{ - 1}} = \frac{1}{{\left( { - 1} \right)\left( 2 \right) - \left( { - 3} \right)\left( 2 \right)}}\left( {\begin{aligned}{*{20}{c}}2&3\\{ - 2}&{ - 1}\end{aligned}} \right)\\ = \frac{1}{4}\left( {\begin{aligned}{*{20}{c}}2&3\\{ - 2}&{ - 1}\end{aligned}} \right)\end{aligned}\)

Therefore, matrix\(A\)is:

\(A = \left( {\begin{aligned}{*{20}{c}}{ - 1}&{ - 3}\\2&2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\0&{.6}\end{aligned}} \right)\frac{1}{4}\left( {\begin{aligned}{*{20}{c}}2&3\\{ - 2}&{ - 1}\end{aligned}} \right)\)

Find the matrix \({A^k}\)

\(\begin{aligned}{c}{A^k} &= \frac{1}{4}\left( {\begin{aligned}{*{20}{c}}{ - 1}&{ - 3}\\2&2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{1^k}}&0\\0&{{{.6}^k}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2&3\\{ - 2}&{ - 1}\end{aligned}} \right)\\ &= \frac{1}{4}\left( {\begin{aligned}{*{20}{c}}{ - 1}&{ - 3}\\2&2\end{aligned}} \right)\left( {\left( {\begin{aligned}{*{20}{c}}{{1^k}}&0\\0&{{{.6}^k}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2&3\\{ - 2}&{ - 1}\end{aligned}} \right)} \right)\\ &= \frac{1}{4}\left( {\begin{aligned}{*{20}{c}}{ - 1}&{ - 3}\\2&2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2&3\\{ - 2{{\left( {.6} \right)}^k}}&{ - {{\left( {.6} \right)}^k}}\end{aligned}} \right)\\ &= \mathop {\lim }\limits_{k \to \infty } \frac{1}{4}\left( {\begin{aligned}{*{20}{c}}{ - 2 + 6{{\left( {.6} \right)}^k}}&{ - 3 + 3{{\left( {.6} \right)}^k}}\\{4 - 4{{\left( {.6} \right)}^k}}&{6 - 2{{\left( {.6} \right)}^k}}\end{aligned}} \right)\\ &= \frac{1}{4}\left( {\begin{aligned}{*{20}{c}}{ - 2}&{ - 3}\\4&6\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{ - .5}&{ - .75}\\1&{1.5}\end{aligned}} \right)\end{aligned}\)

Thus, \(\mathop {\lim }\limits_{k \to \infty } {A^k} \to \left( {\begin{aligned}{*{20}{c}}{ - .5}&{ - .75}\\1&{1.5}\end{aligned}} \right)\).