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I understand that part one is LD because there is a zero vector but part two does not make sense to me

I have seen similar problems on the site but when I apply those methods to my problem, my solution does not match up with the supposed correct answer shown in the image. I know that "If a set is LI and has the same dimensions as the vector space, then it is a basis of V". This set is in fact LI because when I do the method of c1v1 + c2v2 = 0, I get that c1 = c2 = 0 which means LI. Also, when I put the vectors in matrix form and find the rank, I get that the rank is two which matches the dimension of the vector space which indicates, to me, that the set spans P2.

Can someone clarify the method I should be using to approach these problems with polynomials?

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  • $\begingroup$ Your approach seems good to me. Does the program match them as incorrect? $\endgroup$ – Natalio Nov 6 '18 at 0:33
  • $\begingroup$ The dimension of the space is $3.$ Please explain why you think it is $2,$ so we can explain your mistake. $\endgroup$ – saulspatz Nov 6 '18 at 0:36
  • $\begingroup$ @S. Snake, are your answers the ones in the image? I thought they were. $\endgroup$ – Natalio Nov 6 '18 at 0:38
  • $\begingroup$ There are two questions there. Which one are you asking about? $\endgroup$ – amd Nov 6 '18 at 0:43
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0 is in the first set, so it cannot be linearly independent, since $x_10=0$ does not imply $x_1=0$.

The rank of the second set is in fact two, but the dimension of the space is 3. Thus it cannot span $P_2$.

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You know that you have a three dimensional space, so you need three linearly independent vectors in your Basis. If you have four vectors or two vectors it is not a Basis. On the other hand if you have three vectors you may turn them into a matrix and check the determinant for linear independence. Rows of your matrix are coefficients of $\{1,t,t^2\}$ for example $3t^2-2t+1$ turns into $(1,-2,3)$

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