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Do $x^3 − 2x^2 + 1$, $4x^2−x+ 3$ and $3x − 2$ generate the vector space $P_2(ℝ)$; and $P_3(ℝ)$?

My answer was yes, they do generate $P_3(\mathbb{R})$ and thus also $P_2(\mathbb{R})$. I wrote the coefficients of the polynomials as vectors and then added them together using other coefficients, so I could see what the span was. This gave me:

$S=\{(1, 2, 0, 1), (0, 3, -1, 3), (0, 0, 3, -2)\}$ the coefficients of the polynomials written as vectors.

$\text{span}(S)=\{(a, 2a+4b, -b+3c, a+3b-2c):a,b,c\in\mathbb{R}\}$

Let $a=p$, $2a+4b=q$, $-b+3c=r$, $a+3b-2c=s$.

As $p, q, r, s$ are not multiples of each other, the $4$ elements in $\mathbb{R}$ are variable in all 4 dimensions of $S$, so that means the polynomials generate $P_{4-1}(\mathbb{R}) = P_3(\mathbb{R})$. Since $P_2(\mathbb{R})$ ϵ $P_3(\mathbb{R})$, the polynomials also generate $P_2(\mathbb{R})$.

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You are not correct. Note that you can't generate the constant polynomial $1$. In fact the three given polynomials $$x^3 − 2x^2 + 1\;,\;4x^2−x+ 3\;,\;3x − 2$$ have all different degrees greater than zero and therefore any non-zero linear combination has degree greater than $0$.

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  • $\begingroup$ I'm actually really unfamiliar with polynomials and generating; as I can't find a clear example or explanation in my book nor a clear explanation by my professor. I kind of get the idea you're proposing, but I'm struggling to generalize or really understand what is happening here. Would you mind elaborating on the effects of the linear combinations and span of these polynomials? $\endgroup$ – Marc Nov 19 '17 at 14:56

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