# Polynomial Linear dependency

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

• 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. – hardmath Sep 8 '19 at 18:56
• Welcome to Mathematics Stack Exchange. $P_2$ is three-dimensional (spanned by $1, x, x^2$), so certainly four elements are linearly dependent – J. W. Tanner Sep 8 '19 at 19:00
• @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." – Jack Testa Sep 8 '19 at 19:06
• @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. – Jack Testa Sep 8 '19 at 19:08
• @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. – hardmath Sep 8 '19 at 20:25

## 1 Answer

$$P_2$$ 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.

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