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${x^3+x,5x^4-x^3+x^2,3x+2,4x}$ ,

V is the real vector space of all polynomials with real coefficients of degree at most 4.

I know to prove them form a basis, I need to show linear independence and the vectors span V.

I am clear on showing that the vectors are linearly independent, but I am not sure whether I am correct on showing the span or not.

My approach is to set $c_1(x^3+x) + c_2(5x^4 - x^3 + x^2) + c_3 (3x+2) + c_4(4x) = ax^4 + bx^3 + cx^2 + dx^1 + e$ and solve for $c_1,...,c_4$

Then I get $$5c_2 = a$$ $$c_1-c_2 = b$$ $$c_2 = c$$ $$c_1 + 3c_3 + 4c_4 = d$$ $$2c_3 = e$$

And I am confused after this step. From here, do I conclude that since $c_2 = c$ and $c_2 = a/5$, the set of vectors do not span V? My understand is: $a$ and $c$ have to have a 5 to 1 ratio so that the vector space can be written as linear combinations of the set of the vectors given. For example, if $a=5$, and $c=5$, then there does not exist $c_2$, and the vectors do not span V.

Is that correct? Thank you so much!

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Yes your understanding is correct. Note also that you only have $4$ vectors, which can't be enough to span the $5$-dimensional $\Bbb R[X]_4$.

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  • $\begingroup$ Oh that's right! I knew it but I was never really able to visualize it. So in this question, the 5 dimension vector space always have vectors that are out of the span of the set of 4 vectors, and in this case is when a and c do not have a ratio of 5:1. $\endgroup$
    – Valia
    Oct 30, 2016 at 14:23
  • $\begingroup$ @LiseLöwe Exactly, because there is no other obstruction : if the two equations for $c_2$ agree, then you can find $c_1$ (from the second equation), $c_3$ (from the last) and finally $c_4$ (from the fourth). $\endgroup$
    – Arnaud D.
    Oct 30, 2016 at 14:28
  • $\begingroup$ I am sorry this will be my last question. I asked someone else yesterday about the span, and he said since all the terms in V can be found in the set of 4 vectors, the 4 vectors span the vector space. I disagree with him because the definition of span is in this case is that all number of a,b,c,d,e can be written as linear combinations of the 4 vectors, but this is not true because a and c are dependent. Would a easier explanation be because 4 vectors cannot span Dim 5 vector space? $\endgroup$
    – Valia
    Oct 30, 2016 at 14:30

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