Question

[M] Let \(B = \left\{ {{{\bf{x}}_{\bf{0}}},.....,{{\bf{x}}_{\bf{6}}}} \right\}\) and \(C = \left\{ {{{\bf{y}}_{\bf{0}}},.....,{{\bf{y}}_{\bf{6}}}} \right\}\), where \({x_k}\) is the function \({\bf{co}}{{\bf{s}}^k}t\) and \({{\bf{y}}_k}\) is the function \({\bf{cos}}kt\). Exercise 34 in section 4.5 showed that both B and C are basis for the vector space \(H = {\bf{Span}}\left\{ {{{\bf{x}}_{\bf{0}}},....,{{\bf{x}}_{\bf{6}}}} \right\}\).

a. Set \(P = \left[ {{{\left[ {{{\bf{y}}_0}} \right]}_B}\,....\,\,{{\left[ {{{\bf{y}}_{\bf{6}}}} \right]}_B}} \right]\), and calculate \({P^{ - {\bf{1}}}}\).

b. Explain why the columns \({P^{ - {\bf{1}}}}\) are the C-coordinate vectors of \({{\bf{x}}_{\bf{0}}},\,....,\,{{\bf{x}}_{\bf{6}}}\). Then use these coordinate vectors to write trigonometric identities that express powers of \({\bf{cos}}t\) in terms of the functions in C.

Short answer

a. \({P^{ - 1}} = \frac{1}{{32}}\left[ {\begin{array}{*{20}{c}}{32}&0&{16}&0&{12}&0&{10}\\0&{32}&0&{24}&0&{20}&0\\0&0&{16}&0&{16}&0&{15}\\0&0&0&8&0&{10}&0\\0&0&0&0&4&0&6\\0&0&0&0&0&2&0\\0&0&0&0&0&0&1\end{array}} \right]\)

b. Since P is the change of coordinate matrix from C to B, \({P^{ - 1}}\) is the C-coordinate vector of the basis vector in B. The trigonometric identities are listed below:

\(\begin{array}{l}{\cos ^2}t = \frac{1}{2}\left( {1 + \cos 2t} \right)\\{\cos ^3}t = \frac{1}{4}\left( {3\cos t + \cos 3t} \right)\\{\cos ^4}t = \frac{1}{8}\left( {3 + 4\cos 2t + \cos 4t} \right)\\{\cos ^5}t = \frac{1}{{16}}\left( {10\cos t + 5\cos 3t + \cos 5t} \right)\\{\cos ^6}t = \frac{1}{{32}}\left( {10 + 15\cos 2t + 6\cos 4t + \cos 6t} \right)\end{array}\)

Step-by-step solution

Write B-coordinate of each function

The B coordinate of each function can be written as shown below:

\({\left[ {{y_0}} \right]_B} = {\left[ 1 \right]_B} = \left[ {\begin{array}{*{20}{c}}1\\0\\0\\0\\0\\0\\0\end{array}} \right]\), \({\left[ {{y_1}} \right]_B} = {\left[ {\cos t} \right]_s} = \left[ {\begin{array}{*{20}{c}}0\\1\\0\\0\\0\\0\\0\end{array}} \right]\), \({\left[ {{y_2}} \right]_B} = {\left[ {\cos 2t} \right]_S} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\0\\2\\0\\0\\0\\0\end{array}} \right]\), \({\left[ {{y_3}} \right]_B} = {\left[ {\cos 3t} \right]_B} = \left[ {\begin{array}{*{20}{c}}0\\{ - 3}\\0\\4\\0\\0\\0\end{array}} \right]\), \({\left[ {{y_4}} \right]_B} = {\left[ {\cos 4t} \right]_B} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 8}\\0\\8\\0\\0\end{array}} \right]\), \({\left[ {{y_5}} \right]_B} = {\left[ {\cos 5t} \right]_B} = \left[ {\begin{array}{*{20}{c}}0\\5\\0\\{ - 20}\\0\\{16}\\0\end{array}} \right]\), \({\left[ {{y_6}} \right]_B} = {\left[ {\cos 6t} \right]_B} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\0\\{18}\\0\\{ - 48}\\0\\{32}\end{array}} \right]\)

Write matrix P using the B-coordinate

Matrix P is

\(P = \left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&0&1&0&{ - 1}\\0&1&0&{ - 3}&0&5&0\\0&0&2&0&{ - 8}&0&{18}\\0&0&0&4&0&{ - 20}&0\\0&0&0&0&8&0&{ - 48}\\0&0&0&0&0&{16}&0\\0&0&0&0&0&0&{32}\end{array}} \right]\).

Find the inverse of P

Use the following code in MATLAB to find theinverse of P:

\(\begin{array}{l} > > P = \left[ \begin{array}{l}\begin{array}{*{20}{c}}1&0&{ - 1}&0&1&0&{ - 1;\,}\end{array}\,\,\begin{array}{*{20}{c}}0&1&0&{ - 3}&0&5&{0;\,}\end{array}\,\begin{array}{*{20}{c}}0&0&2&0&{ - 8}&0&{18;\,}\end{array}\\\begin{array}{*{20}{c}}0&0&0&4&0&{ - 20}&{0;\,\,\begin{array}{*{20}{c}}0&0&0&0&8&0&{ - 48;\,\,\begin{array}{*{20}{c}}0&0&0&0&0&{16}&{0;\,}\end{array}}\end{array}}\end{array}\\\begin{array}{*{20}{c}}0&0&{ - 1}&0&1&0&{ - 1;\,}\end{array}\end{array} \right];\\ > > X = {\rm{inv}}\left( P \right)\end{array}\)

So,

\({P^{ - 1}} = \frac{1}{{32}}\left[ {\begin{array}{*{20}{c}}{32}&0&{16}&0&{12}&0&{10}\\0&{32}&0&{24}&0&{20}&0\\0&0&{16}&0&{16}&0&{15}\\0&0&0&8&0&{10}&0\\0&0&0&0&4&0&6\\0&0&0&0&0&2&0\\0&0&0&0&0&0&1\end{array}} \right]\)

Write the coordinate vectors using trigonometric identities

Since P is the change of coordinate matrix from C to B, \({P^{ - 1}}\) is the C-coordinate vector of the basis vector in B. The trigonometric identities are listed below:

\(\begin{array}{l}{\cos ^2}t = \frac{1}{2}\left( {1 + \cos 2t} \right)\\{\cos ^3}t = \frac{1}{4}\left( {3\cos t + \cos 3t} \right)\\{\cos ^4}t = \frac{1}{8}\left( {3 + 4\cos 2t + \cos 4t} \right)\\{\cos ^5}t = \frac{1}{{16}}\left( {10\cos t + 5\cos 3t + \cos 5t} \right)\\{\cos ^6}t = \frac{1}{{32}}\left( {10 + 15\cos 2t + 6\cos 4t + \cos 6t} \right)\end{array}\)

So, the inverse of P is \(\frac{1}{{32}}\left[ {\begin{array}{*{20}{c}}{32}&0&{16}&0&{12}&0&{10}\\0&{32}&0&{24}&0&{20}&0\\0&0&{16}&0&{16}&0&{15}\\0&0&0&8&0&{10}&0\\0&0&0&0&4&0&6\\0&0&0&0&0&2&0\\0&0&0&0&0&0&1\end{array}} \right]\).