Let \({{\bf{p}}_1}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = t - {\bf{3}}{t^{\bf{2}}}\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{1}} + t - {\bf{3}}{t^{\bf{2}}}\).

a. Use coordinate vectors to show that these polynomials form a basis of \({{\bf{P}}_{\bf{2}}}\).

b. Consider the basis \(B = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}}} \right\}\) for \({{\bf{P}}_{\bf{2}}}\). Find \({\bf{q}}\) in \({{\bf{P}}_{\bf{2}}}\), given that \({\left( {\bf{q}} \right)_B} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\)

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

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

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

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

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

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

The required vector can be calculated as follows:

\(\begin{array}{c}{\bf{q}} = {P_B}{\left( {\bf{q}} \right)_B}\\ = \left( {{{\bf{p}}_1},\;{{\bf{p}}_2},\;{{\bf{p}}_3}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}\\1\\2\end{array}} \right)\\ = \left( { - 1} \right)\left( {1 + {t^2}} \right) + \left( 1 \right)\left( {t - 3{t^2}} \right) + \left( 2 \right)\left( {1 + t - 3{t^2}} \right)\\ = 1 + 3t - 10{t^2}\end{array}\)

Therefore, the required vector is \({\bf{q}} = 1 + 3t - 10{t^2}\).