Question

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

Short answer

The set $\left\{p_1,p_2\right\}$ is a basis for $\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\left\{p_1,p_2,p_3\right\}$.

Step-by-step solution

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

$\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\left\{p_1,p_2,p_3\right\}=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\left\{p_1,p_2\right\}$.

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.

Draw a conclusion

Hence, $\left\{p_1,p_2\right\}$ forms a basis for $\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\left\{p_1,p_2,p_3\right\}$.