# Show that the set $S = \{1, x+1 , x^2+x+1\}$ generates $P_2(R)$

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.

• Very similar: math.stackexchange.com/questions/951481/… May 15, 2016 at 18:44
• @AritraDas you are wrong, this is not at all similar to that problem. Not even the same type of mathematics involved. May 16, 2016 at 1:08
• 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. May 16, 2016 at 4:42

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.

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.

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

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

• 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. May 15, 2016 at 18:54

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 $s'$.