Question

In Exercises 33 and 34, determine whether the sets of polynomials form a basis for \({{\bf{P}}_3}\). Justify your conclusions.

(M) \({\bf{3}} + {\bf{7}}t\), \({\bf{5}} + t - {\bf{2}}{t^{\bf{3}}}\), \(t - {\bf{2}}{t^{\bf{2}}}\), \({\bf{1}} + {\bf{16}}t - {\bf{6}}{t^{\bf{2}}} + {\bf{2}}{t^{\bf{3}}}\)

Short answer

Set S is not the basis for \({{\bf{P}}_3}\).

Step-by-step solution

Write the polynomials in the standard vector form

\(\left\{ {3 + 7t,\;5 + t - 2{t^3},\;t - 2{t^2},\;1 + 16t - 6{t^2} + 2{t^3}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}3\\7\\0\\0\end{array}} \right),\;\;\left( {\begin{array}{*{20}{c}}5\\1\\0\\{ - 2}\end{array}} \right),\;\;\left( {\begin{array}{*{20}{c}}0\\1\\{ - 2}\\0\end{array}} \right),\;\;\left( {\begin{array}{*{20}{c}}1\\{16}\\{ - 6}\\2\end{array}} \right)} \right\}\)

Form the matrix using the vectors

The matrix formed by using the vectors of the polynomials is:

\(A = \left( {\begin{array}{*{20}{c}}3&5&0&1\\7&1&1&{16}\\0&0&{ - 2}&{ - 6}\\0&{ - 2}&0&2\end{array}} \right)\)

Write the matrix in the echelon form

Consider matrix \(A = \left( {\begin{array}{*{20}{c}}3&5&0&1\\7&1&1&{16}\\0&0&{ - 2}&{ - 6}\\0&{ - 2}&0&2\end{array}} \right)\).

Use code in the MATLAB to obtain the row-reducedechelon form as shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left( {{\rm{ }}\begin{array}{*{20}{c}}3&5&0&{1;\;\;\begin{array}{*{20}{c}}7&1&1&{16;\;\;\begin{array}{*{20}{c}}0&0&{ - 2}&{ - 6;\;\;\begin{array}{*{20}{c}}0&{ - 2}&0&2\end{array}}\end{array}}\end{array}}\end{array}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left( {\begin{array}{*{20}{c}}3&5&0&1\\7&1&1&{16}\\0&0&{ - 2}&{ - 6}\\0&{ - 2}&0&2\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&0&2\\0&1&0&{ - 1}\\0&0&1&3\\0&0&0&0\end{array}} \right)\)

It can be observed from the echelon form that the last row is zero, which shows the set is linearly dependent.

Therefore, set S is not the basis for \({{\bf{P}}_3}\).