Question

Let A be a \(2 \times 2\) matrix with eigenvalues \(3\) and \(1/3\) and corresponding eigenvectors \({{\rm{v}}_1} = \left( {\begin{aligned}{}1\\1\end{aligned}} \right)\) and \({{\rm{v}}_2} = \left( {\begin{aligned}{}{ - 1}\\1\end{aligned}} \right)\).Let \(\left\{ {{{\rm{x}}_k}} \right\}\) be a solution of the difference equation \({{\rm{x}}_{k + 1}} = A{{\rm{x}}_k},{{\rm{x}}_0} = \left( {\begin{aligned}{}9\\1\end{aligned}} \right)\) .

1. Compute\({{\rm{x}}_1} = A{{\rm{x}}_0}\). (Hint: You do not need to know A itself.
2. Find a formula for\({{\rm{x}}_k}\)involving k and the eigenvectors\({{\rm{v}}_1}\)and\({{\rm{v}}_2}\).

Short answer

(a)The solution is \({{\rm{x}}_1} = \left( {\begin{aligned}{}{49/3}\\{41/3}\end{aligned}} \right)\).

(a)The formula for \({{\rm{x}}_k}\)is \({{\rm{x}}_k} = 5{\left( 3 \right)^k}{{\rm{v}}_1} - 4{\left( {1/3} \right)^k}{{\rm{v}}_2}\)

Step-by-step solution

Write the condition for the vector and a matrix 

a)

Express \({x_0}\) in terms of \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\) to find A's action on\({x_0}\) . To put it another way, find \({c_1}\) and \({c_2}\) such that \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2}\). Because the eigenvectors \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\) are linearly independent (by inspection and also because they correspond to unique eigenvalues), this is a distinct possibility.

\({R^2}\) has a foundation (In \({R^2}\), two linearly independent vectors span \({R^2}\) automatically.)

Row reduction in matrix

A row can be reduced with help of a vector\({{\rm{v}}_1}\) and \({{\rm{v}}_2}\).

Shows that \({{\rm{x}}_0} = 5{{\rm{v}}_1} - 4{{\rm{v}}_2}\).

Since \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\). are the Eigenvector for Eigenvalue \(3\) and\({{\rm{x}}_k} = 5{\left( 3 \right)^k}{{\rm{v}}_1} - 4{\left( {1/3} \right)^k}{{\rm{v}}_2},k \ge 0\).

\({x_1} = A{x_0}\)

Put the value of \({x_0}\)in the above equation.

\({{\rm{x}}_1} = 5A{{\rm{v}}_1} - 4A{{\rm{v}}_2}\)

Put the eigenvalue in the above equation.

\({{\rm{x}}_1} = 5.3{{\rm{v}}_1} - 4.\frac{1}{3}{{\rm{v}}_2}\)

Put all the values in the above equation.

\(\begin{aligned}{}{{\rm{x}}_1} &= \left( {\begin{aligned}{}{15}\\{15}\end{aligned}} \right) - \left( {\begin{aligned}{}{ - 4/3}\\{4/3}\end{aligned}} \right)\\{{\rm{x}}_1} &= \left( {\begin{aligned}{}{49/3}\\{41/3}\end{aligned}} \right)\end{aligned}\)

Write the condition for the vector and a matrix 

b)

The \({{\rm{v}}_1}\) term is multiplied by the eigenvalue 3 and the \({{\rm{v}}_2}\) term is multiplied by the eigenvalue 1/3 each time A operates on a linear combination of \({{\rm{v}}_1}\) and \({{\rm{v}}_2}\):

Then the value of \({{\rm{x}}_2}\) is \({{\rm{x}}_2} = A{{\rm{x}}_1}\).

Put the value of \({{\rm{x}}_{\rm{1}}}\) in the above equation

\(\begin{aligned}{}{{\rm{x}}_2} &= A\left( {5.3{{\rm{v}}_1} - 4\left( {1/3} \right){{\rm{v}}_2}} \right)\\{{\rm{x}}_2} &= 5{\left( 3 \right)^2}{{\rm{v}}_1} - 4{\left( {1/3} \right)^2}{{\rm{v}}_2}\end{aligned}\)

In the general form, we can write

\({{\rm{x}}_k} = 5{\left( 3 \right)^k}{{\rm{v}}_1} - 4{\left( {1/3} \right)^k}{{\rm{v}}_2}\)for \(k \ge 0\).