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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

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  • $\begingroup$ 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
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    $\begingroup$ @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
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$P_2$ 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.

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

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