Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\), \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\), \({{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\), \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\) and \({\rm{w}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

1. Show that \({{\rm{v}}_1}\), \({{\rm{v}}_2}\), and \({{\rm{v}}_3}\) are eigenvectors of \(A\). (Note: \(A\) is the stochastic matrix studied in Example 3 of Section 4.9.)
2. Let \({{\rm{x}}_0}\) be any vector in \({\mathbb{R}^3}\) with non-negative entries whose sum is 1. (In section 4.9, \({{\rm{x}}_0}\) was called a probability vector.) Explain why there are constants \({c_1}\), \({c_2}\), and \({c_3}\) such that \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\). Compute \({{\rm{w}}^T}{{\rm{x}}_0}\), and deduce that \({c_1} = 1\).
3. For \(k = 1,2, \ldots ,\) define \({{\rm{x}}_k} = {A^k}{{\rm{x}}_0}\), with \({{\rm{x}}_0}\) as in part (b). Show that \({{\rm{x}}_k} \to {{\rm{v}}_1}\) as \(k\) increases.

Find the value of \(A{{\rm{v}}_1}\), \(A{{\rm{v}}_2}\) and \(A{{\rm{v}}_3}\).

\[\begin{array}A{{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\\ = {{\rm{v}}_1}\end{array}\]

\[\begin{array}A{{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.5}\\{ - 1.5}\\1\end{array}} \right)\\ = 0.5\left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\\ = 0.5{{\rm{v}}_2}\end{array}\]

And,

\[\begin{array}{c}A{{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\\ = 0.2{{\rm{v}}_3}\end{array}\]

As \(A{{\rm{v}}_1} = {{\rm{v}}_1}\), \(A{{\rm{v}}_2} = 0.5{{\rm{v}}_2}\) and \(A{{\rm{v}}_3} = 0.2{{\rm{v}}_3}\), this implies that \({{\rm{v}}_1}\), \({{\rm{v}}_2}\) and \({{\rm{v}}_3}\) are eigenvectors of \(A\).

If \({v_1}, \ldots ,{v_r}\) are eigenvectors that correspond to distinct eigenvalues \({\lambda _1}, \ldots ,{\lambda _r}\) of an \(n \times n\) matrix \(A\), then the set \(\left\{ {{v_1}, \ldots ,{v_r}} \right\}\) is linearly independent.

This implies that \(\left\{ {{{\rm{v}}_1},{{\rm{v}}_2},{{\rm{v}}_3}} \right\}\) is linearly independent, and the vectors in the set are the basis for \({\mathbb{R}^3}\).

There exists a constant which satisfies the condition \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\).

Write \({{\rm{w}}^T}{{\rm{x}}_0}\) as:

\[{{\rm{w}}^T}{{\rm{x}}_0} = {c_1}{{\rm{w}}^T}{{\rm{v}}_1} + {c_2}{{\rm{w}}^T}{{\rm{v}}_2} + {c_3}{{\rm{w}}^T}{{\rm{v}}_3}\]

As \({{\rm{x}}_0}\) and \({{\rm{v}}_1}\) are probability vectors and the sum of the entries of vector \({{\rm{v}}_2}\) and \({{\rm{v}}_3}\) is zero, so from the obtained equation, we obtain that \({c_1} = 1\).

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

Substitute \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\) into \({x_k} = {A^k}{x_0}\) and simplify.

\[\begin{array}{c}{x_k} = {A^k}\left( {{c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}} \right)\\ = {c_1}{A^k}{{\rm{v}}_1} + {c_2}{A^k}{{\rm{v}}_2} + {c_3}{A^k}{{\rm{v}}_3}\\ = {{\rm{v}}_1} + {c_2}{\left( {0.5} \right)^k}{{\rm{v}}_2} + {c_3}{\left( {0.2} \right)^k}{{\rm{v}}_3}\end{array}\]

It is observed that as the value of \(k\) increases, \({x_k} \to {v_1}\).

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