Question

(M) Show thatis a linearly independent set of functions defined on. Use the method of Exercise 37. (This result will be needed in Exercise 34 in Section 4.5.)

Short answer

By the invertible matrix theorem, is a linearly independent set of functions defined on .

Step-by-step solution

Use the method used in Exercise 37

Assume \({c_1} \cdot 1 + {c_2} \cdot \cos t + {c_3} \cdot {\cos ^2}t + ... + {c_7} \cdot {\cos ^6}t = 0\).

For \(t = 0,.1,.2,...,.6\), the above equation gives the system as shown below:

\(\left( {\begin{array}{*{20}{c}}1&{\cos 0}&{{{\cos }^2}0}&{{{\cos }^3}0}&{{{\cos }^4}0}&{{{\cos }^5}0}&{{{\cos }^6}0}\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\\{{c_4}}\\{{c_5}}\\{{c_6}}\\{{c_7}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\\0\\0\\0\\0\end{array}} \right)\)

That is, \(Ac = 0\).

Find the determinant of A

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&{\cos 0}&{{{\cos }^2}0}&{{{\cos }^3}0}&{{{\cos }^4}0}&{{{\cos }^5}0}&{{{\cos }^6}0}\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}1&1&1&1&1&1&1\\1&{\cos .1}&{{{\cos }^2}.1}&{{{\cos }^3}.1}&{{{\cos }^4}.1}&{{{\cos }^5}.1}&{{{\cos }^6}.1}\\1&{\cos .2}&{{{\cos }^2}.2}&{{{\cos }^3}.2}&{{{\cos }^4}.2}&{{{\cos }^5}.2}&{{{\cos }^6}.2}\\1&{\cos .3}&{{{\cos }^2}.3}&{{{\cos }^3}.3}&{{{\cos }^4}.3}&{{{\cos }^5}.3}&{{{\cos }^6}.3}\\1&{\cos .4}&{{{\cos }^2}.4}&{{{\cos }^3}.4}&{{{\cos }^4}.4}&{{{\cos }^5}.4}&{{{\cos }^6}.4}\\1&{\cos .5}&{{{\cos }^2}.5}&{{{\cos }^3}.5}&{{{\cos }^4}.5}&{{{\cos }^5}.5}&{{{\cos }^6}.5}\\1&{\cos .6}&{{{\cos }^2}.6}&{{{\cos }^3}.6}&{{{\cos }^4}.6}&{{{\cos }^5}.6}&{{{\cos }^6}.6}\end{array}} \right|\\\det A \ne 0\end{array}\)

Draw a conclusion

By the invertible matrix theorem, has only a trivial solution. Thus, is a linearly independent set of functions defined on .