Question

15.Let \(A = \left( {\begin{aligned}{}{.4}&{}&0&{}&{.2}\\{.3}&{}&{.8}&{}&{.3}\\{.3}&{}&{.2}&{}&{.5}\end{aligned}} \right)\). The vector \({v_1} = \left( {\begin{aligned}{}{.1}\\{.6}\\{.3}\end{aligned}} \right)\) is an eigenvector for \(A\), and two eigenvalues are .5 and .2. Construct the solution of the dynamical system \({{\bf{x}}_{k + 1}} = A{{\bf{x}}_k}\) that satisfies \({x_0} = \left( {0,\,\,.3,\,\,.7} \right)\). What happens to \({x_k}\) as \(k \to \infty \)?

Short answer

The solution to the dynamical system is:

\({x_k} = {v_1} + \left( {0.1} \right){\left( {0.5} \right)^k}{v_2} + \left( {0.3} \right){\left( {0.2} \right)^k}{v_3}\)

And, \({x_k}\) approach to \({v_1}\) as \(k \to \infty \).

Step-by-step solution

Discrete Dynamical System 

For anyDynamical Systemwith matrix \(A\), the eigenvalues of \(A\) will determine the system of:

\(\begin{aligned}{}{\rm{Greatest attraction }} \to {\rm{if }}\lambda < 1\\{\rm{Greatest repulsion }} \to {\rm{if }}\lambda > 1\end{aligned}\)

Evaluating Eigenvalue of given eigenvector

The given matrix and the eigenvector for the dynamic system \({x_{k + 1}} = A{x_k}\) are respectively:

\(\begin{aligned}{}A = \left( {\begin{aligned}{}{.4}&{}&0&{}&{.2}\\{.3}&{}&{.8}&{}&{.3}\\{.3}&{}&{.2}&{}&{.5}\end{aligned}} \right)\\{v_1} = \left( {\begin{aligned}{}{.1}\\{.6}\\{.3}\end{aligned}} \right)\end{aligned}\)

The two eigenvalues are given, that is, 0.5 and 0.2. Now, we have:

\(\begin{aligned}{c}A{v_1} &= \left( {\begin{aligned}{}{.4}&{}&0&{}&{.2}\\{.3}&{}&{.8}&{}&{.3}\\{.3}&{}&{.2}&{}&{.5}\end{aligned}} \right)\left( {\begin{aligned}{}{.1}\\{.6}\\{.3}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{.1}\\{.6}\\{.3}\end{aligned}} \right)\\& = 1.{v_1}\end{aligned}\)

So, the eigenvalue found is \(\lambda = 1\).

Eigenvector for eigenvalue 0.5 

Now, for eigenvectors, we have:

At \(\lambda = 0.5\);

\(\begin{aligned}{}\left( {A - \left( {0.5} \right)I} \right){v_2} &= 0\\\left( {\begin{aligned}{}{0.4 - 0.5}&{}&0&{}&{0.2}\\{0.3}&{}&{0.8 - 0.5}&{}&{0.3}\\{0.3}&{}&{0.2}&{}&{0.5 - 0.5}\end{aligned}} \right)\left( {\begin{aligned}{}x\\\begin{aligned}{}y\\z\end{aligned}\end{aligned}} \right) &= 0\\\left( {\begin{aligned}{}{ - 0.1}&{}&0&{}&{0.2}\\{0.3}&{}&{0.3}&{}&{0.3}\\{0.3}&{}&{}&{0.2}&{}&0\end{aligned}} \right)\left( {\begin{aligned}{}x\\\begin{aligned}{l}y\\z\end{aligned}\end{aligned}} \right) &= 0\end{aligned}\)

Also, the system of equationwill be:

\(\begin{aligned}{c} - 0.1x + 0.2z = 0\\0.3x + 0.3y + 0.3z = 0\\0.3x + 0.2y = 0\end{aligned}\)

For this, the general solution, we have:

\(\begin{aligned}{}{v_2} &= \left( {\begin{aligned}{}x\\\begin{aligned}{l}y\\z\end{aligned}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}2\\\begin{aligned}{l} - 3\\1\end{aligned}\end{aligned}} \right)\end{aligned}\)

Here, the eigenvector is: \({v_2} = \left( {\begin{aligned}{}2\\\begin{aligned}{l} - 3\\1\end{aligned}\end{aligned}} \right)\).

Eigenvector for eigenvalue 0.2

Now, for eigenvectors, we have:

At \(\lambda = 0.2\).

\(\begin{aligned}{}\left( {A - \left( {0.2} \right)I} \right){v_3} &= 0\\\left( {\begin{aligned}{}{0.4 - 0.2}&{}&0&{}&{0.2}\\{0.3}&{}&{0.8 - 0.2}&{}&{0.3}\\{0.3}&{}&{0.2}&{}&{0.5 - 0.2}\end{aligned}} \right)\left( {\begin{aligned}{}x\\\begin{aligned}{l}y\\z\end{aligned}\end{aligned}} \right) &= 0\\\left( {\begin{aligned}{}{0.2}&{}&0&{}&{0.2}\\{0.3}&{}&{0.6}&{}&{0.3}\\{0.3}&{}&{0.2}&{}&{0.3}\end{aligned}} \right)\left( {\begin{aligned}{}x\\\begin{aligned}{l}y\\z\end{aligned}\end{aligned}} \right) &= 0\end{aligned}\)

Also, the system of equationwill be:

\(\begin{aligned}{}0.2x + 0.2z = 0\\0.3x + 0.6y + 0.3z = 0\\0.3x + 0.2y + 0.3z = 0\end{aligned}\)

For this, the general solution, we have:

\(\begin{aligned}{}{v_3} &= \left( {\begin{aligned}{}x\\\begin{aligned}{l}y\\z\end{aligned}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - 1}\\\begin{aligned}{l}0\\1\end{aligned}\end{aligned}} \right)\end{aligned}\)

Here, the eigenvector is \({v_3} = \left( {\begin{aligned}{}{ - 1}\\\begin{aligned}{}0\\1\end{aligned}\end{aligned}} \right)\).

Constructing the Solution of the system

Now, according to the question, we have:

\({x_0} = \left( {0,\,\,.3,\,\,.7} \right)\)

We know:

\({x_0} = {c_1}{v_1} + {c_2}{v_2} + {c_3}{v_3}\)

So,

\(\begin{aligned}{}\left( {\begin{aligned}{}{{v_1}}&{{v_2}}&{{v_3}}&{{x_0}}\end{aligned}} \right) &= \left( {\begin{aligned}{}{0.1}&{}&2&{}&{ - 1}&{}&0\\{0.6}&{}&{ - 3}&{}&0&{}&{0.3}\\{0.3}&{}&1&{}&1&{}&{0.7}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}1&{}&0&{}&0&{}&1\\0&{}&1&{}&0&{}&{0.1}\\0&{}&0&{}&1&{}&{0.3}\end{aligned}} \right)\end{aligned}\)

Thus, the general solution can be:

\(\begin{aligned}{}{x_0} = {v_1} + 0.1{v_2} + 0.3{v_3}\\{x_1} &= A{v_1} + 0.1A{v_2} + 0.3A{v_3}\\ = {v_1} + 0.1\left( {0.5} \right){v_2} + 0.3\left( {0.2} \right){v_3}\\{x_k} &= {v_1} + \left( {0.1} \right){\left( {0.5} \right)^k}{v_2} + \left( {0.3} \right){\left( {0.2} \right)^k}{v_3}\end{aligned}\)

Here, we see, \({x_k}\) is approaching to \({v_1}\) as the value of \(k\) is increasing.

Hence, the solution to the dynamical system is: \({x_k} = {v_1} + \left( {0.1} \right){\left( {0.5} \right)^k}{v_2} + \left( {0.3} \right){\left( {0.2} \right)^k}{v_3}\) and \({x_k}\) approach to \({v_1}\) as \(k \to \infty \).