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Let $V = \mathbb R[x]_{\le 3}$

I have the space of polynomials $U_2 = \{ p = a_0 + a_1x + a_2x^2 + a_3x^3 \in V \mid a_1 - a_2 + a_3 = 0, a_0 = a_1 \}$

I am asked to find a basis, so I proceed by noticing that in $U_2$:

$$a_0 + a_1x + a_2x^2 + a_3x^3 = a_0 (1+x-x^3) + a_2 (x^2 +x^3)$$

So I figure that a basis is $(1+x-x^3,x^2 + x^3 )$ In $\mathbb R^4$ the vectors would be $(1 10 -1), (001 1)$ but the solution gives the vectors $(1110), (-1-1 0 1)$.

What am I doing wrong?

Edit: corrected an error pointed out in the comments.

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  • $\begingroup$ $x^2-x^3$ has $a_0=a_1=0,a_2=1,a_3=-1$, so $a_1-a_2+a_3=-2\ne0$. $\endgroup$ – almagest Jun 21 '16 at 16:04
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    $\begingroup$ should be $a_2(x^2+x^3)$, check again! $\endgroup$ – Euler_Salter Jun 21 '16 at 16:04
  • $\begingroup$ I don't understand how you have written your vectors $\endgroup$ – Euler_Salter Jun 21 '16 at 16:09
  • $\begingroup$ @Euler_Salter I'm probably wrong, how should I write them? I was thinking the first component represents the coefficient of the $x^0$, the second the coefficient of the $x^1$ and so on... $\endgroup$ – Monolite Jun 21 '16 at 16:14
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    $\begingroup$ just with commas! $\endgroup$ – Euler_Salter Jun 21 '16 at 16:26
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$a_2=a_1+a_3$ $\therefore $ the space is spanned by $a_0(1+x+x^2)+a_3(x^2+x_3)$ So you get $(1,1,1,0)$ and $(0,0,1,1)$.

You can also write it as $a_0(1+x-x^3)+a_2(x^2+x^3)$ In this case you get your answer which is also absolutely correct. Remember their is no unique solution to this question.

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These are both bases. Your basis is $\{1+x-x^3, x^2+x^3\}$; the solution gives the basis $\{1+x+x^2, -1-x+x^3\}$. But each of these is expressible in terms of the other: \begin{gather*} 1+x+x^2 = (1+x-x^3) + (x^2+x^3),\quad -1-x+x^3 = -(1+x-x^3) \\ 1+x-x^3 = -(-1-x+x^3),\quad x^2+x^3 = (1+x+x^2) + (-1-x+x^3). \end{gather*}

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