Decide if the following polynomials in $P_2$ are linearly dependent. If so, write one polynomial as a linear combination of the others.

$p_1 = 1+x+x^2, \ \ \ p_2 = 7+2x, \ \ \ p_3 = -1+5x^2, \ \ \ p_4 = 6-7x^2$

Solution I have equated them to 0 to get the following linear system:$$\alpha_1+7\alpha_2-\alpha_3+6\alpha_4=0$$ $$\alpha_1+2\alpha_2=0$$ $$\alpha_1+5\alpha_3-7\alpha_4$$

I put them in an augmented matrix and got the following R.E.F: $\begin{bmatrix}1 & 7 & -1&6&0\\0 & -7&6&-13&0 \\ 0&0&\frac{-23}7&\frac{23}7&0 \end{bmatrix}$

I am not sure how to proceed further please help me

  • $\begingroup$ This is a good approach. Now one of your unknowns in the row echelon form does not correspond to a leading one in the system. So assign a nonzero value to it and work out the values of the other variables. $\endgroup$ – hardmath Sep 8 '19 at 18:56
  • $\begingroup$ Welcome to Mathematics Stack Exchange. $P_2$ is three-dimensional (spanned by $1, x, x^2$), so certainly four elements are linearly dependent $\endgroup$ – J. W. Tanner Sep 8 '19 at 19:00
  • $\begingroup$ @hardmath thank you. I have the following: $\alpha_ 1 = 2\alpha_4, \alpha_2 = -\alpha_4$ and $\alpha_3 =\alpha_4$. How do I "write one polynomial as a linear combination of the others." $\endgroup$ – Jack Testa Sep 8 '19 at 19:06
  • $\begingroup$ @J.W.Tanner thank you. that is a good shortcut you have mentioned. I would make my assumption and work on from there in the future. $\endgroup$ – Jack Testa Sep 8 '19 at 19:08
  • 1
    $\begingroup$ @JackTesta: Setting $\alpha_4 = 1$ gives (in effect) an expression for $p_4$ as a linear combination of the other polynomials. While some shortcuts can be seen to work in this problem, it's a good exercise to show how a "hammer and tongs" approach produces the same solution. $\endgroup$ – hardmath Sep 8 '19 at 20:25

$P_2$ is three-dimensional (spanned by $1,x,x^2$), so certainly four elements are linearly dependent, and so we should be able to express $p_4$ as a linear combination of $p_1, p_2, p_3$. Note that $p_4$ and $p_3$ have no $x$ term. Furthermore, $p_2-2p_1=5-2x^2$ is a linear combination of $p_1$ and $p_2$ with no $x$ term. It is then easy to see that $p_2-2p_1-p_3=(5-2x^2)-(-1+5x^2)=6-7x^2=p_4,$ so that's the answer.

  • $\begingroup$ thanks a lot. sorry for over complicating things $\endgroup$ – Jack Testa Sep 8 '19 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.