Question

Suppose the eigenvalue a\(3 \times 3\) matrix A are\(3\),\(4/5\)and\(3/5\)with corresponding eigenvectors \(\left[ {\begin{array}{*{20}{c}}1\\0\\{ - 3}\end{array}} \right]\)\(\left[ {\begin{array}{*{20}{c}}2\\1\\{ - 5}\end{array}} \right]\) and\(\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 3}\\7\end{array}} \right]\).Let \[{x_0} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 5}\\3\end{array}} \right]\] . Find the solution \({x_{k + 1}} = A{x_k}\) .For the specified \[{x_0}\]and describe what happen at\(x \to \infty \)

Short answer

The general solution is \({{\rm{x}}_k} \approx 2 \cdot {3^k},{{\rm{v}}_1} = 2 \cdot {3^k}\left( {\begin{aligned}{{}}1\\0\\{ - 3}\end{aligned}} \right)\).

Step-by-step solution

Given term for the vector and a matrix

The vectors are \({{\rm{v}}_1} = \left( {\begin{aligned}{{}}1\\0\\{ - 3}\end{aligned}} \right),{{\rm{v}}_2} = \left( {\begin{aligned}{{}}2\\1\\{ - 5}\end{aligned}} \right),{{\rm{v}}_3} = \left( {\begin{aligned}{{}}{ - 3}\\{ - 3}\\7\end{aligned}} \right)\).

These are the eigenvectors of \(3 \times 3\)matrices A.

3, 4/5, and 3/5 are the eigenvalues of matrices A and\({{\rm{x}}_0} = \left( {\begin{aligned}{{}}{ - 2}\\{ - 5}\\3\end{aligned}} \right)\).

Solution of the equation of matrix 

The solution of the equation of matrix

\({{\rm{x}}_{k + 1}} = A{{\rm{x}}_k}\left( {k = 1,2,....} \right),\)

First,write \({x_0}\) in terms of eigenvectors

\({x_0} = 2{v_1} + {v_2} + 2{v_3}\)

\({{\rm{x}}_0} = 2{{\rm{v}}_1} + {{\rm{v}}_2} + 2{{\rm{v}}_3}\)

Then,

\(\begin{aligned}{}{{\rm{x}}_1} = A\left( {2{{\rm{v}}_1} + {{\rm{v}}_2} + 2{{\rm{v}}_3}} \right)\\{{\rm{x}}_1} = 2A{{\rm{v}}_1} + A{{\rm{v}}_2} + 2A{{\rm{v}}_3}\end{aligned}\)

Put the eigenvalue in the above equation

\(\begin{aligned}{}{{\rm{x}}_1} = 2 \cdot 3{{\rm{v}}_1} + \left( {4/5} \right){{\rm{v}}_2} + 2 \cdot \left( {3/5} \right){{\rm{v}}_3}\\{{\rm{x}}_k} = 2 \cdot {3^k}{{\rm{v}}_1} + {\left( {4/5} \right)^k}{{\rm{v}}_2} + 2 \cdot {\left( {3/5} \right)^k}{{\rm{v}}_3}\\{{\rm{x}}_k} \approx 2 \cdot {3^k}{{\rm{v}}_1}\\ = 2 \cdot {3^k}\left( {\begin{aligned}{{}}1\\0\\{ - 3}\end{aligned}} \right)\end{aligned}\)

The general solution is \({{\rm{x}}_k} \approx 2 \cdot {3^k}{{\rm{v}}_1}\).

For all k sufficiently large

Then,

\(\begin{aligned}{}{x_k} \approx 2 \cdot {3^k}{{\rm{v}}_1}\\ = 2 \cdot {3^k}\left( {\begin{aligned}{{}}1\\0\\{ - 3}\end{aligned}} \right)\end{aligned}\).

This is the final solution.