Question

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

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

Step-by-step solution

Write the linear combination

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

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

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

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

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

Draw a conclusion

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