Question

Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)

Short answer

The basis of H is \(\left\{ {{{\bf{p}}_1},\;{{\bf{p}}_2},\,{{\bf{p}}_4}} \right\}\).

Step-by-step solution

Check for a combination of \({{\bf{p}}_{\bf{1}}}\) and \({{\bf{p}}_{\bf{2}}}\)

For a scalar \({c_1}\),

\(\begin{array}{c}{p_2} = {c_1}{p_1}\\1 - t = {c_1}\left( {1 + t} \right)\end{array}\).

The above equation is not true for any value of \({c_1}\). So, \({p_1}\) and \({p_2}\) are independent.

Check for a linear combination of \({{\bf{p}}_{\bf{3}}}\)

For scalars \({c_1}\) and \({c_2}\),

\(\begin{aligned}{c}{{\bf{p}}_3} &= {c_1}{{\bf{p}}_1} + {c_2}{{\bf{p}}_2}\\4 &= {c_1}\left( {1 + t} \right) + {c_2}\left( {1 - t} \right)\\ &= {c_1} + {c_2} + \left( {{c_1} - {c_2}} \right)t\end{aligned}\).

For the above equation, \({c_1} = 2\) and \({c_2} = 2\).

So, \({{\bf{p}}_3}\) can be obtained from \({{\bf{p}}_1}\) and \({{\bf{p}}_2}\).

Check for a linear combination of \({{\bf{p}}_{\bf{4}}}\)

For the scalars \({c_1}\), \({c_2}\), and \({c_3}\),

\(\begin{aligned} {{\bf{p}}_4} &= {c_1}{{\bf{p}}_1} + {c_2}{{\bf{p}}_2} + {c_3}{{\bf{p}}_3}\\t + {t^2} &= {c_1}\left( {1 + t} \right) + {c_2}\left( {1 - t} \right) + {c_3}\left( 4 \right)\\ &= {c_1} + {c_2} + 4{c_3} + \left( {{c_1} - {c_2}} \right)t\end{aligned}\).

As \({t^2}\) is not present in RHS, \({p_4}\) is an independent work.

Check for a linear combination of \({{\bf{p}}_{\bf{5}}}\)

For the scalars \({c_1}\), \({c_2}\), \({c_3}\), and \({c_4}\),

\(\begin{aligned}{c}{{\bf{p}}_5} &= {c_1}{{\bf{p}}_1} + {c_2}{{\bf{p}}_2} + {c_3}{{\bf{p}}_3} + {c_4}{{\bf{p}}_4}\\1 + 2t + {t^2} &= {c_1}\left( {1 + t} \right) + {c_2}\left( {1 - t} \right) + {c_3}\left( 4 \right) + {c_4}\left( {t + {t^2}} \right)\\ &= {c_1} + {c_2} + 4{c_3} + \left( {{c_1} - {c_2} + {c_4}} \right){t^2}\end{aligned}\).

For the above equation, \({c_1} = 1\), \({c_2} = 0\), \({c_3} = 0\), and \({c_4} = 1\).

So, \({p_5}\) is a dependent work.

The basis is formed by the independent vectors, so the basis of H is \(\left\{ {{{\bf{p}}_1},\;{{\bf{p}}_2},\,{{\bf{p}}_4}} \right\}\).