Suggested languages for you:

Americas

Europe

Q33E

Expert-verifiedFound in: Page 191

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Consider the polynomials , and \({p_{\bf{3}}}\left( t \right) = {\bf{2}}\) \({p_{\bf{1}}}\left( t \right) = {\bf{1}} + t,{p_{\bf{2}}}\left( t \right) = {\bf{1}} - t\)(for all t). By inspection, write a linear dependence relation among \({p_{\bf{1}}},{p_{\bf{2}}},\) and \({p_{\bf{3}}}\). Then find a basis for Span\(\left\{ {{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\).**

The set \(\left\{ {{p_1},{p_2}} \right\}\) is a basis for \({\rm{Span}}\left\{ {{p_1},{p_2},{p_3}} \right\}\).

\(\begin{array}{c}{p_1}\left( t \right) + {p_2}\left( t \right) = 1 + t + 1 - t\\ = 2\\{p_1}\left( t \right) + {p_2}\left( t \right) = {p_3}\left( t \right)\end{array}\)

By **the spanning set theorem**, you get

\({\rm{Span}}\left\{ {{p_1},{p_2},{p_3}} \right\} = {\rm{Span}}\left\{ {{p_1},{p_2}} \right\}\).

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

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

This implies,

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

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

This implies \(\left\{ {{p_1},{p_2}} \right\}\) is **linearly independent**.

Hence, \(\left\{ {{p_1},{p_2}} \right\}\) forms a basis for \({\rm{Span}}\left\{ {{p_1},{p_2},{p_3}} \right\}\).

94% of StudySmarter users get better grades.

Sign up for free