Answer to Question #311136 in Linear Algebra for Jlove

Question

Which of the following sets of polynomials span P2 ?

a. {t²+1, t-1, t²+t}

B. {t²+ 2, 2t²- t+1, t + 2 , t²+t+4}

Answer 1

Solution (a)

 

If \left\{ {{t^2} + 1,\,\,t - 1,\,\,{t^2} + t} \right\}\ , span P2, then {t^2} + 1,\,\,t - 1,\,\,{t^2} + t\ must be linearly independent.

Then for, three arbitrary constants ${c_1},\,\,{c_2},\,\,{c_3}$

$\begin{array}{l} {c_1}\left( {{t^2} + 1} \right) + {c_2}\left( {t - 1} \right) + {c_3}\left( {{t^2} + t} \right) = 0\\ {c_1}{t^2} + {c_1} + {c_2}t - {c_2} + {c_3}{t^2} + {c_3}t = 0\\ \left( {{c_1} + {c_3}} \right){t^2} + {c_2}t + {c_1} - {c_2} = 0 \cdot {t^2} + 0 \cdot t + 0 \end{array}$

This gives,

${c_1} + {c_3} = 0\\ {c_2} = 0\\ {c_1} - {c_2} = 0$

We solve and get {c_1} = {c_2} = {c_3} = 0\

Therefore,  {t^2} + 1,\,\,t - 1,\,\,{t^2} + t\are linearly independent and hence span polynomial of degree 2. 

Solution (b)

If, \left\{ {{t^2} + 2,\,\,\,2{t^2} - t + 1,\,\,\,t + 2,\,\,{t^2} + t + 4} \right\}\ span P2, then ${{t^2} + 2,\,\,\,2{t^2} - t + 1,\,\,\,t + 2,\,\,{t^2} + t + 4}$ must be linearly independent.

Then for, four arbitrary constants {c_1},\,\,{c_2},\,\,{c_3},\,{c_4}\

$\begin{array}{l} {t^2} + 2,\,\,\,2{t^2} - t + 1,\,\,\,t + 2,\,\,{t^2} + t + 4\\ {c_1}\left( {{t^2} + 2} \right) + {c_2}\left( {2{t^2} - t + 1} \right) + {c_3}\left( {t + 2} \right) + {c_4}\left( {{t^2} + t + 4} \right) = 0\\ {c_1}{t^2} + 2{c_1} + 2{c_2}{t^2} - {c_2}t + {c_2} + {c_3}t + 2{c_3} + {c_4}{t^2} + {c_4}t + 4{c_4} = 0\\ \left( {{c_1} + 2{c_2} + {c_4}} \right){t^2} + \left( { - {c_2} + {c_3} + {c_4}} \right)t + \left( {2{c_1} + {c_2} + 2{c_3} + 4{c_4}} \right) = 0 \cdot {t^2} + 0 \cdot t + 0 \end{array}$

This gives,

$\begin{array}{l} {c_1} + 2{c_2} + {c_4} = 0 & & ... & \left( 1 \right)\\ - {c_2} + {c_3} + {c_4} = 0 & & ... & \left( 2 \right)\\ 2{c_1} + {c_2} + 2{c_3} + 4{c_4} & & ... & \left( 3 \right) \end{array}$

These are three equations, and we have four unknowns. So, there are infinite solutions.

Therefore, the are linearly dependent and hence ${{t^2} + 2,\,\,\,2{t^2} - t + 1,\,\,\,t + 2,\,\,{t^2} + t + 4}$are linearly dependent and hence does not span polynomial of degree 2.