Show that the signals in Exercises 3-6 form a basis for the solution set of the accompanying difference equation.\(\left\{ {{3^k},\,\,{{\left( { - 3} \right)}^k}} \right\}\)

The signals and equation in Exercise 2.

For scalars \({c_1}\) and \({c_2}\), \({c_1}{\left( 3 \right)^k} + {c_2}{\left( { - 3} \right)^k} = 0\).

Write the above equation in the matrix form.

\(\begin{aligned} \left[ {\begin{array}{*{20}{c}}{{3^k}}&{{{\left( { - 3} \right)}^k}}\\{{3^{k + 1}}}&{{{\left( { - 3} \right)}^{k + 1}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\\{A_k}c &= 0\end{aligned}\)

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

For \(k = 0\),

\(\begin{aligned} \left[ {\begin{array}{*{20}{c}}{{3^k}}&{{{\left( { - 3} \right)}^k}}\\{{3^{k + 1}}}&{{{\left( { - 3} \right)}^{k + 1}}}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{{3^0}}&{{{\left( { - 3} \right)}^0}}\\{{3^1}}&{{{\left( { - 3} \right)}^1}}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}1&1\\3&{ - 3}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}1&1\\0&{ - 6}\end{array}} \right].\end{aligned}\)

There is a pivot column in each row, and the system has atrivial solution. So, the signals \(\left\{ {{3^k},\,\,{{\left( { - 3} \right)}^k}} \right\}\) are linearly independent.

The solution set of the equation \({y_{k + 2}} - 9{y_k} = 0\) is two-dimensional andlinearly independent.

Therefore, by the basis theorem, the set \(\left\{ {{3^k},\,\,{{\left( { - 3} \right)}^k}} \right\}\) forms a basis for the difference equation \({y_{k + 2}} - 9{y_k} = 0\).