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Show that the set $S = \{1, x+1 , x^2+x+1\}$ generates $p_2(R)$. I am stuck as to how to show this. Linear concepts are not making sense to me. Any help would be appreciated. Thank you.

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  • $\begingroup$ Very similar: math.stackexchange.com/questions/951481/… $\endgroup$
    – Aritra Das
    May 15, 2016 at 18:44
  • $\begingroup$ @AritraDas you are wrong, this is not at all similar to that problem. Not even the same type of mathematics involved. $\endgroup$ May 16, 2016 at 1:08
  • $\begingroup$ It is the exact same mathematics (linear algebra) and almost the same problem. Even some of the answers on this post are the same as the ones in my link. $\endgroup$
    – Aritra Das
    May 16, 2016 at 4:42

4 Answers 4

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If you show $s$ is linearly independent, that is if $\lambda_1, \lambda_2, \lambda_3 \in \mathbb{R}$ and $$\lambda_1 + \lambda_2(x+1) +\lambda_3(x^2+x+1) = 0$$ for all $x \in \mathbb{R}$ then $\lambda_1 = \lambda_2 = \lambda_3 = 0$, then $s$ is a set of 3 linearly independent vectors in a 3 dimensional vector space. Now a basic theorem of linear algebra says that this implies $s$ is a basis.

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There are some very good answers, but I thought I'd add this:

Consider the equivalence $$ 1 \to (1\;0\;0)\\ (1+x) \to (1\;1\;0) \\ (1+x+x^2)\to (1\;1\;1) $$ Then we can simply check that the determinant $$ \det\pmatrix{1&0&0\\1&1&0\\1&1&1}$$ is non-zero hence the claim is confirmed.

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To show that $s$ spans $\mathbb{R}_{\leq 2}[x]$, you need to take a general second degree polynomial $p(x) = a_0 + a_1x + a_2x^2$ and show that $p$ can be written as a linear combination of the members of the set $s$. Namely, you need to show that you can find $b_0, b_1, b_2 \in \mathbb{R}$ such that

$$ p(x) = a_0 + a_1x + a_2x^2 = b_0 \cdot 1 + b_1 \cdot (x + 1) + b_2 \cdot (x^2 + x + 1). $$

By expanding the right hand side and collecting terms, we get

$$ a_0 + a_1x + a_2x^2 = (b_0 + b_1 + b_2) \cdot 1 + (b_1 + b_2) \cdot x + b_2 \cdot x^2. $$

Comparing coefficients, we see that

$$ b_2 = a_2, \\ b_1 + b_2 = a_1 \implies b_1 = a_2 - b_2 = a_1 - a_2, \\ b_0 + b_1 + b_2 = a_0 \implies b_0 = a_0 - b_1 - b_2 = a_0 - (a_1 - a_2) - a_2 = a_0 - a_1. $$

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    $\begingroup$ If you have difficulty understanding, try to show how $2+4x-3x^2$ can be written as $b_0 \cdot 1 + b_1 \cdot (x + 1) + b_2 \cdot (x^2 + x + 1)$. And then try $21+34x-53x^2$. You could, of course in an infinite length of time, prove that all polynomials of degree 2 or less can be written as a combination of $\{1,x+1,x^2+x+1\}$, but you can show this in one calculation given above. $\endgroup$ May 15, 2016 at 18:54
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If you can start from the assumption that set $s' = \{1, x, x^2\}$ is a generator for it, then you can simply observe that:

$$ 1 = 1 \\ x = (x + 1) + (-1) \cdot 1 \\ x^2 = (x^2 + x + 1) + (-1) \cdot (x + 1) $$

So, you can generate $s'$ from $s$, which means that you can generate with $s$ whatever can be generated with $s'$.

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