Question

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.6}&{.3}\\{.4}&{.7}\end{array}} \right)\), \({v_1} = \left( {\begin{array}{*{20}{c}}{3/7}\\{4/7}\end{array}} \right)\), \({x_0} = \left( {\begin{array}{*{20}{c}}{.5}\\{.5}\end{array}} \right)\). (Note: \(A\) is the stochastic matrix studied in Example 5 of Section 4.9.)

1. Find a basic for \({\mathbb{R}^2}\) consisting of \({{\rm{v}}_1}\) and anther eigenvector \({{\rm{v}}_2}\) of \(A\).
2. Verify that \({{\rm{x}}_0}\) may be written in the form \({{\rm{x}}_0} = {{\rm{v}}_1} + c{{\rm{v}}_2}\).
3. For \(k = 1,2, \ldots \), define \({x_k} = {A^k}{x_0}\). Compute \({x_1}\) and \({x_2}\), and write a formula for \({x_k}\). Then show that \({{\bf{x}}_k} \to {{\bf{v}}_1}\) as \(k\) increases.

Short answer

1. The basis for \({\mathbb{R}^2}\) is \(\left\{ {{v_1},{v_2}} \right\}\).
2. It is verified that \({{\bf{x}}_0}\) can be written in the form \({{\bf{x}}_0} = {{\bf{v}}_1} + c{{\bf{v}}_2}\).
3. \({{\bf{x}}_1} = {{\bf{v}}_1} + c\left( {0.3} \right){{\bf{v}}_2}\), \({{\bf{x}}_2} = {{\rm{v}}_1} + c\left( {0.3} \right)\left( {0.3} \right){{\rm{v}}_2}\), \({{\bf{x}}_k} = {{\bf{v}}_1} + c{\left( {0.3} \right)^k}{{\bf{v}}_2}\). It is proved that \({{\bf{x}}_k} \to {{\bf{v}}_1}\) as \(k\) increases.

Step-by-step solution

(a) Step 1: Find a basic for \({\mathbb{R}^2}\) consisting of \({{\rm{v}}_1}\) and another eigenvector \({{\rm{v}}_2}\) of \(A\)

It is given that \(A\) be a \(2 \times 2\) matrix. Find the eigenvalues of the matrix using the characteristic polynomial.

\[\begin{array}{c}\det \left( {A - \lambda I} \right) = 0\\\det \left( {\left( {\begin{array}{*{20}{c}}{.6 - \lambda }&{.3}\\{.4}&{.7 - \lambda }\end{array}} \right)} \right) = 0\\\left( {.6 - \lambda } \right)\left( {.7 - \lambda } \right) - \left( {.3} \right)\left( {.4} \right) = 0\\.42 - 0.6\lambda - .7\lambda + {\lambda ^2} - .12 = 0\\{\lambda ^2} - 1.3\lambda + .3 = 0\\\left( {\lambda - 1} \right)\left( {\lambda - .3} \right) = 0\\\lambda = 1,.3\end{array}\]

This implies that the eigenvalues are \(1\) and \(.3\).

For the eigenvalue \(\lambda = .3\), Solve the equation \(\left( {A - .3I} \right)x = 0\).

\[\begin{array}{c}\left( {\begin{array}{*{20}{c}}{.6 - .3}&{.3}&0\\{.4}&{.7 - .3}&0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{.3}&{.3}&0\\{.4}&{.4}&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&1&0\\0&0&0\end{array}} \right)\end{array}\]

It is observed that \({x_1} + {x_2} = 0\) with \({x_2}\) is free. Consider \({x_2} = 1\) and find the eigenvector \({v_2}\).

\[{v_2} = \left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\]

This implies that basis for \({\mathbb{R}^2}\) is \(\left\{ {{v_1},{v_2}} \right\}\).

(b) Step 2: Verify that \({{\rm{x}}_0}\) may be written in the form \({{\rm{x}}_0} = {{\rm{v}}_1} + c{{\rm{v}}_2}\)

Substitute the given values into the equation \({x_0} = {v_1} + c{v_2}\).

\({x_0} = {v_1} + c{v_2}\)

\[\begin{array}{c}\left( {\begin{array}{*{20}{c}}{.5}\\{.5}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{3/7}\\{4/7}\end{array}} \right) + {\rm{c}}\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{\frac{1}{2}}\\{\frac{1}{2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{3}{7} - {\rm{c}}}\\{\frac{4}{7} + {\rm{c}}}\end{array}} \right)\\{\rm{c}} = \frac{{ - 1}}{{14}}\end{array}\]

This implies that \({x_0}\) can be written in the form \({x_0} = {v_1} + c{v_2}\).

(c) Step 3: Compute \({x_1}\) and \({x_2}\), and write a formula for \({x_k}\) also show that \({x_k} \to {v_1}\) as \(k\) increases

Consider \(k = 1,2,3, \ldots \) , and it is given that \({x_k} = {A^k}{x_0}\).

Substitute \(k = 1\) into \({x_k} = {A^k}{x_0}\) and simplify to write \({x_1}\). Use the fact that \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\) are eigenvectors.

\(\begin{array}{c}{x_1} = {A^1}{x_0}\\ = A\left( {{{\rm{v}}_1} + c{{\rm{v}}_2}} \right)\\ = A{{\rm{v}}_1} + cA{{\rm{v}}_2}\\ = {{\rm{v}}_1} + c\left( {0.3} \right){{\rm{v}}_2}\end{array}\)

Similarly, write \({x_2}\).

\[\begin{array}{c}{x_2} = A{x_1}\\ = A\left( {{{\rm{v}}_1} + c\left( {0.3} \right){{\rm{v}}_2}} \right)\\ = A{{\rm{v}}_1} + c\left( {0.3} \right)A{{\rm{v}}_2}\\ = {{\rm{v}}_1} + c\left( {0.3} \right)\left( {0.3} \right){{\rm{v}}_2}\end{array}\]

It is observed that, \({x_k} = {v_1} + c{\left( {0.3} \right)^k}{v_2}\).

It is observed that as the value of \(k\) increases, the second term \(c{\left( {0.3} \right)^k}{v_2}\) tend to be 0. This implies that as \(k\) increases, \({x_k} \to {v_1}\).

It is proved that \({x_k} \to {v_1}\) as \(k\) increases.