Question

Show that the signals in Exercises 3-6 form a basis for the solution set of the accompanying difference equation.

The signals and equation in Exercise 1.

Short answer

\(\left\{ {{2^k},{{\left( { - 4} \right)}^k}} \right\}\) forms a basis for the difference equation.

Step-by-step solution

Check the dependence of the solution

For any scalar \({c_1}\) and \({c_2}\),

\({c_1}{\left( 2 \right)^k} + {c_2}{\left( { - 4} \right)^k} = 0\).

Write the above equation in the matrix form.

\(\begin{aligned} \left[ {\begin{array}{*{20}{c}}{{2^k}}&{{{\left( { - 4} \right)}^k}}\\{{2^{k + 1}}}&{{{\left( { - 4} \right)}^{k + 1}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\\{A_k}c &= 0\end{aligned}\)

So, \({3^k}\) is the solution of the given difference equation.

Write the augmented matrix for the matrix equation

For \(k = 0\), you get

\(\begin{aligned} \left[ {\begin{array}{*{20}{c}}{{2^k}}&{{{\left( { - 4} \right)}^k}}\\{{2^{k + 1}}}&{{{\left( { - 4} \right)}^{k + 1}}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{{2^0}}&{{{\left( { - 4} \right)}^0}}\\{{2^1}}&{{{\left( { - 4} \right)}^1}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}1&1\\2&{ - 4}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}1&1\\0&{ - 6}\end{array}} \right].\end{aligned}\)

There is a pivot column in each row, and the system has a trivial solution. So, the signals \(\left\{ {{2^k},\,\,{{\left( { - 4} \right)}^k}} \right\}\) are linearly independent.

Check whether the solution is the basis of the difference equation

The solution set of the equation \({y_{k + 2}} + 2{y_{k + 1}} - 8{y_k} = 0\) is two dimensional and linearly independent.

Therefore, by the basis theorem, the set \(\left\{ {{2^k},{{\left( { - 4} \right)}^k}} \right\}\) forms a basis for the difference equation \({y_{k + 2}} + 2{y_{k + 1}} - 8{y_k} = 0\).