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Linear Algebra and its Applications
Found in: Page 191
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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

See the step by step solution

Step by Step Solution

Step 1: Find the relation among the polynomials

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

By the spanning set theorem, you get

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

Step 2: Use the definition of linear independence

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.

Step 3: Draw a conclusion

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

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