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Consider the polynomials \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}},{p_{\bf{2}}}\left( t \right) = {\bf{1}} - {t^{\bf{2}}}\). Is \(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}}} \right\}\) a linearly independent set in \({{\bf{P}}_{\bf{3}}}\)? Why or why not?

Short Answer

Expert verified

The set \(\left\{ {{p_1},{p_2}} \right\}\) is linearly independent in \({{\rm{P}}_3}\).

Step by step solution

01

Write the linear combination

Consider the linear combination of \({p_1}\) and \({p_2}\).

\(\begin{array}{c}{c_1}{p_1}\left( t \right) + {c_2}{p_2}\left( t \right) = 0\\{c_1}\left( {1 + {t^2}} \right) + {c_2}\left( {1 - {t^2}} \right) = 0\\{c_1} + {c_1}{t^2} + {c_2} - {c_2}{t^2} = 0\\\left( {{c_1} + {c_2}} \right) + \left( {{c_1} - {c_2}} \right){t^2} = 0\end{array}\)

02

Solve \({c_{\bf{1}}}\) and \({c_{\bf{2}}}\)

\(\begin{array}{l}{c_1} + {c_2} = 0\\{c_1} - {c_2} = 0\end{array}\)

Thus, \({c_1} = {c_2} = 0\).

03

Draw a conclusion

This implies, \(\left\{ {{p_1},{p_2}} \right\}\) is linearly independent in \({{\rm{P}}_3}\).

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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.)

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