Question

Question: Describe how might try to build a solution of a difference equation \({{\rm{x}}_{k + 1}} = A{{\rm{x}}_k}\,\,\,\,\left( {k = 0,1,2,...} \right)\) if you were given the initial \({{\rm{x}}_0}\) and this vector did not happen to be an eigenvector of A.

Short answer

The solution of a difference equation \(A{x_k} = {x_{k + 1}}\) can be built by using the linear combination of eigenvectors.

Step-by-step solution

Write \({{\rm{x}}_0}\) as a linear combination of eigenvectors

Here, \({x_0}\) can be written as a linear combination of eigenvectors as follows:

\({x_0} = {c_1}{v_1} + ... + {c_p}{v_p}\)

Computation of \(A{x_k}\)

Hence, we can define the recursive description of the sequence \(\left\{ {{x_k}} \right\}\), using the above definition, as

\({x_k} = {c_1}{\lambda ^k}{v_1} + ... + {c_p}{\lambda ^k}{v_p}\) for \(k = 0,1,2,...\)

Replace \(k\) by \(k + 1\) in the above equation to get:

\({x_{k + 1}} = {c_1}{\lambda ^{k + 1}}{v_1} + ... + {c_p}{\lambda ^{k + 1}}{v_p}\)

Now, compute \(A{x_k}\) as follows:

\(\begin{array}{c}A{x_k} = A\left( {{c_1}{\lambda ^k}{v_1} + ... + {c_p}{\lambda ^k}{v_p}} \right)\\ = {c_1}{\lambda ^k}A{v_1} + ... + {c_p}{\lambda ^k}A{v_p}\,\,\,{\rm{(using linearity property)}}\\ = {c_1}{\lambda ^k}\lambda {v_1} + ... + {c_p}{\lambda ^k}\lambda {v_p}\,\,({\rm{using definition of eigenvectors}})\\ = {c_1}{\lambda ^{k + 1}}{v_1} + ... + {c_p}{\lambda ^{k + 1}}{v_p}\\ = {x_{k + 1}}\end{array}\)

Hence, \(A{x_k} = {x_{k + 1}}\).