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

\({{\bf{5}}^k}{\bf{cos}}\frac{{k\pi }}{{\bf{2}}}\), \({{\bf{5}}^k}sin\frac{{k\pi }}{{\bf{2}}}\), \({y_{k + {\bf{2}}}} + {\bf{25}}{y_k} = {\bf{0}}\)

Substitute\({y_k} = {5^k}\cos \frac{{k\pi }}{2}\)in the difference equation\({y_{k + 2}} + 25{y_k} = 0\).

\(\begin{aligned} {y_{k + 2}} + 25{y_k} &= {5^{k + 2}}\cos \frac{{\left( {k + 2} \right)\pi }}{2} + 25\left( {{5^k}\cos \frac{{k\pi }}{2}} \right)\\ &= {5^k}\left( {{5^2}\cos \frac{{\left( {k + 2} \right)\pi }}{2} + 25\cos \frac{{k\pi }}{2}} \right)\\ &= 25 \cdot {5^k}\left( {\cos \left( {\frac{{k\pi }}{2} + \pi } \right) + \cos \frac{{k\pi }}{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\cos \left( {t + \pi } \right) = - \cos t} \right]\\ &= 25 \cdot {5^k}\left( 0 \right)\end{aligned}\)

Substitute\({y_k} = {5^k}\sin \frac{{k\pi }}{2}\)in the difference equation\({y_{k + 2}} + 25{y_k} = 0\).

\(\begin{aligned} {y_{k + 2}} + 25{y_k} &= {5^{k + 2}}\sin \frac{{\left( {k + 2} \right)\pi }}{2} + 25\left( {{5^k}\sin \frac{{k\pi }}{2}} \right)\\ &= {5^k}\left( {{5^2}\sin \frac{{\left( {k + 2} \right)\pi }}{2} + 25\sin \frac{{k\pi }}{2}} \right)\\ &= 25 \cdot {5^k}\left( {\sin \left( {\frac{{k\pi }}{2} + \pi } \right) + \sin \frac{{k\pi }}{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\sin \left( {t + \pi } \right) = - \sin t} \right]\\ &= 25 \cdot {5^k}\left( 0 \right)\end{aligned}\)

For all k, in n-dimensional vector space, the dimension of H is 2. So,\({5^k}\cos \frac{{k\pi }}{2}\)and\({5^k}\sin \frac{{k\pi }}{2}\)form the basis of the solution setfor the difference equation.