Question

Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.

a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)

b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)

Short answer

a. The set does not span for \({P_2}\).

b. The set spans for \({P_2}\).

Step-by-step solution

Write the polynomials in the standard vector form

\(\left\{ {1 - 3t + 5{t^2}, - 3 + 5t - 7{t^2}, - 4 + 5t - 6{t^2},1 - {t^2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}1\\{ - 3}\\5\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - 3}\\5\\{ - 7}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - 4}\\5\\{ - 6}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right)} \right\}\)

Form the matrix using the vectors

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

\(A = \left( {\begin{array}{*{20}{c}}1&{ - 3}&{ - 4}&1\\{ - 3}&5&5&0\\5&{ - 7}&{ - 6}&{ - 1}\end{array}} \right)\)

Write the matrix in the echelon form

\(\left( {\begin{array}{*{20}{c}}1&{ - 3}&{ - 4}&1\\{ - 3}&5&5&0\\5&{ - 7}&{ - 6}&{ - 1}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&{\frac{5}{4}}&{ - \frac{5}{4}}\\0&1&{\frac{7}{4}}&{ - \frac{3}{4}}\\0&0&0&0\end{array}} \right)\)

From the echelon form, it can be observed that for three variables, there are two equations. Hence, there is onefree variable.

So, the given set does not span for \({P_2}\).

Write the polynomials in the  standard vector form

\(\left\{ {5t + {t^2},1 - 8t - 2{t^2}, - 3 + 4t + 2{t^2},2 - 3t} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}0\\5\\1\end{array}} \right),\,\,\left( {\begin{array}{*{20}{c}}1\\{ - 8}\\{ - 2}\end{array}} \right),\,\,\left( {\begin{array}{*{20}{c}}{ - 3}\\4\\2\end{array}} \right),\,\,\left( {\begin{array}{*{20}{c}}2\\{ - 3}\\0\end{array}} \right)} \right\}\)

Form the matrix using the vectors

\(B = \left( {\begin{array}{*{20}{c}}0&1&{ - 3}&2\\5&{ - 8}&4&{ - 3}\\1&{ - 2}&2&0\end{array}} \right)\)

Write the matrix in the echelon form

\(\left( {\begin{array}{*{20}{c}}0&1&{ - 3}&2\\5&{ - 8}&4&{ - 3}\\1&{ - 2}&2&0\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&0&{ - 4}&0\\0&1&{ - 3}&0\\0&0&0&1\end{array}} \right)\)

From the echelon form, it can be observed that for three variables, there are three equations. Hence, no free variable is present.

Therefore, the set spans for \({P_2}\).