Question

[M] Let \[B = \left\{ {{\bf{1}},{\bf{cos}}t,{\bf{co}}{{\bf{s}}^{\bf{2}}}t,....,{\bf{co}}{{\bf{s}}^{\bf{6}}}t} \right\}\] and \(C = \left\{ {{\bf{1}},{\bf{cos}}t, {\bf{cos2}}t,......{\bf{cos6}}t} \right\}\). Assume the following trigonometric identities (see Exercises 37 in section 4.1).

\[{\bf{cos2}}t = - {\bf{1}} + {\bf{2co}}{{\bf{s}}^{\bf{2}}}t\]

\[{\bf{cos3}}t = - {\bf{3cos}}t + {\bf{4co}}{{\bf{s}}^{\bf{3}}}t\]

\[{\bf{cos4}}t = {\bf{1}} - {\bf{8co}}{{\bf{s}}^{\bf{2}}}t + {\bf{8co}}{{\bf{s}}^{\bf{4}}}t\]

\[{\bf{cos5}}t = {\bf{5cos}}t - {\bf{20co}}{{\bf{s}}^{\bf{3}}}t + {\bf{16co}}{{\bf{s}}^{\bf{5}}}t\]

\[{\bf{cos6}}t = - {\bf{1}} + {\bf{18co}}{{\bf{s}}^{\bf{2}}}t - {\bf{48co}}{{\bf{s}}^{\bf{4}}}t + {\bf{32co}}{{\bf{s}}^{\bf{6}}}t\]

Let H be the subspace of functions spanned by the functions in B. Then B is a basis for H, by Exercises 38 in section 4.3.

a. Write the B-coordinte vectors of the vectors in C, and use them to show that C is a linearly independent set in H.

b. Explain why C is a basis for H.

Short answer

a. \[\left[ {\begin{array}{*{20}{c}}{{b_0}}&{{b_1}}&{{b_2}}&{{b_3}}& \ldots &{{b_6}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&0&1&0&{ - 1}\\0&1&0&{ - 3}&0&5&0\\0&0&2&0&{ - 8}&0&{18}\\0&0&0&4&0&{ - 20}&0\\0&0&0&0&8&0&{ - 48}\\0&0&0&0&0&{16}&0\\0&0&0&0&0&0&{32}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\{\cos t}\\{{{\cos }^2}t}\\{{{\cos }^3}t}\\{{{\cos }^4}t}\\{{{\cos }^5}t}\\{{{\cos }^6}t}\end{array}} \right] = 0\]

b. The vectors in C are linearly independent. Therefore, C is the basis of H.

Step-by-step solution

Write the B coordinate of each vector in C

Let

\({\left[ 1 \right]_B} = \left[ {\begin{array}{*{20}{c}}1\\0\\0\\0\\0\\0\\0\end{array}} \right]\)

Then,

\({\left[ 1 \right]_B} = \left[ {1\,\,\cos t\,\,{{\cos }^2}t\,\,{{\cos }^3}t\,\,.....\,\,{{\cos }^6}t} \right]\)

And

\(\begin{array}{l}{\left[ {\cos t} \right]_B} = \left[ {\begin{array}{*{20}{c}}0\\1\\0\\0\\0\\0\\0\end{array}} \right]\\\cos t = {\left[ {\cos t} \right]_s}\left[ {1\;\;\cos t\;\;{{\cos }^2}t\;\;{{\cos }^3}t\;\;....\;\;{{\cos }^6}t} \right]\end{array}\)

Write the vectors for the given trigonometric identities

For \(\cos 2t = - 1 + 2{\cos ^2}t\):

\(\cos 2t = {\left[ {\cos 2t} \right]_B}\left[ {1\;\;\cos t\;\;{{\cos }^2}t\,\,{{\cos }^3}t\,....\,\,{{\cos }^6}t} \right]\)

For \(\cos 3t = - 3\cos t + 4{\cos ^3}t\):

\(\cos 3t = {\left[ {\cos 3t} \right]_B}\left[ {1\;\;\cos t\;\;{{\cos }^2}t\,\,{{\cos }^3}t\,....\,\,{{\cos }^6}t} \right]\)

For \(\cos 4t = 1 - 8{\cos ^2}t + 8{\cos ^4}t\):

\(\cos 4t = {\left[ {\cos 4t} \right]_B}\left[ {1\;\;\cos t\;\;{{\cos }^2}t\,\,{{\cos }^3}t\,....\,\,{{\cos }^6}t} \right]\)

For \(\cos 5t = 5\cos t - 20{\cos ^3}t + 16{\cos ^5}t\):

\(\cos 5t = {\left[ {\cos 5t} \right]_B}\left[ {1\;\;\cos t\;\;{{\cos }^2}t\,\,{{\cos }^3}t\,....\,\,{{\cos }^6}t} \right]\)

For \(\cos 6t = - 1 + 18{\cos ^2}t - 48{\cos ^4}t + 32{\cos ^6}t\):

\(\cos 6t = {\left[ {\cos 6t} \right]_B}\left[ {1\;\;\cos t\;\;{{\cos }^2}t\,\,{{\cos }^3}t\,....\,\,{{\cos }^6}t} \right]\)

Write the vectors of trigonometric identity in the matrix form

C is a linearly independent set in H if

\({b_0}\left( 1 \right) + {b_1}\cos t + {b_2}\cos 2t + .... + {b_6}\cos 6t = 0\)

The matrix form of the equation is:

\[\left[ {\begin{array}{*{20}{c}}{{b_0}}&{{b_1}}&{{b_2}}&{{b_3}}& \ldots &{{b_6}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&0&1&0&{ - 1}\\0&1&0&{ - 3}&0&5&0\\0&0&2&0&{ - 8}&0&{18}\\0&0&0&4&0&{ - 20}&0\\0&0&0&0&8&0&{ - 48}\\0&0&0&0&0&{16}&0\\0&0&0&0&0&0&{32}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\{\cos t}\\{{{\cos }^2}t}\\{{{\cos }^3}t}\\{{{\cos }^4}t}\\{{{\cos }^5}t}\\{{{\cos }^6}t}\end{array}} \right] = 0\]

Check why Col A\( = {\mathbb{R}^n}\)

Let

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

Enter matrix A.

\(\begin{array}{l} > > A = \left[ \begin{array}{l}1\,\,0\,\, - 1\,\,0\,\,1\,\,0\,\, - 1\,;\,\,0\,\,1\,\,0\,\, - 3\,\,0\,\,0\,\,5\,\,0\,;\,\,0\,\,0\,\,2\,\,0\,\, - 8\,\,0\,\,18;\,\,0\,\,0\,\,0\,\,4\,\,\,0\,\,\, - 20\,\,\,0;\,\,0\,\,0\,\,0\,\,0\,\,8\,\,0\,\, - 48;\,\,\\0\,\,0\,\,0\,\,0\,\,0\,\,16\,\,0;\,\,0\,\,0\,\,0\,\,0\,\,0\,\,0\,\,32\end{array} \right];\\ > > U = {\rm{null}}\left( A \right)\end{array}\)

The null matrix is a zero matrix. Therefore, set C is linearly independent in subspace H, and the dimension of subspace H is 7.

So, set C is the basis of space H.