I'm having trouble understanding the answer given. So I solved it down and found the rref of the matrix, but I don't udnerstand why my solution isn't right.
Consider a set of polynomials $S = \{p_1,p_2,p_3,p_4\}$ in $P_2$, where

  • $p_1(z) = 1+z-z^2$,
  • $p_2(z) = 2-z$,
  • $p_3(z) = 5-4z + z^2$,
  • $p_4(z) = z^2$.
Show that $S$ is a linearly dependent spanning set for $P_2$, and then find a subset of $S$ which is a basis for $P_2$.

So what I did was found the span of the $p_k$'s and removed one polynomial. However, the one I removed was $p_4$.
The answer says a basis for $P_2$ would be $\{p_1,p_2,p_4\},\{p_2,p_3,p_4\},\{p_1,p_3,p_4\}$ NOT $\{p_1,p_2,p_3\}$.
I don't understand, because when I find the linear combination of $p_1,p_2,p_3$, it is independent and spans $P_2$ as well? Why is my answer not correct?


  • 1
    $\begingroup$ How to MathJax: to get $P_n$ you type $P_n$. To get $W = \{ (x,y,z) \in \Bbb F \mid x=y^2=2z \}$ you type $W = \{ (x,y,z) \in \Bbb F \mid x=y^2=2z \}$. To get $$A\mathbf x = \mathrm b$$ you type $$A\mathbf x = \mathrm b$$. Now see if you can edit your question to make it more readable. $\endgroup$ – user137731 Oct 6 '16 at 13:36
  • $\begingroup$ Sorry about that! Will use that from now on. Also thank you to Sloan for proposing the edit to me as well $\endgroup$ – OneGapLater Oct 6 '16 at 13:41
  • $\begingroup$ @ActuarialStudent101: How is proof that $p_1, p_2, p_3$ are linear independent? $\endgroup$ – Moritz Oct 6 '16 at 13:44
  • $\begingroup$ Hmm, I must've made a really stupid mistake. Would there be an easy way to know which polynomial to remove when doing the 3x4 matrix (in this case)? $\endgroup$ – OneGapLater Oct 6 '16 at 13:51
  • $\begingroup$ Reduce row with coefficients of polynomial in columns. Then find a column without a pivot. That corresponds to a polynomial that can be removed. $\endgroup$ – user137731 Oct 6 '16 at 13:52

You can easily check that:

$ p_1+p_3 = -3z+6 = 3 (p_2) $

So the set $ \{p_1,p_2,p_3\}$ is not a basis because it is not linearly independent.

  • $\begingroup$ Thanks! Wouldn't that mean the basis should not have either $p_1$ or $p_2$? Since a $p_2$ can be formed by a linear combination of the other polynomials? $\endgroup$ – OneGapLater Oct 6 '16 at 13:49
  • $\begingroup$ It means that $\{p_1,p_2,p_3\}$ is not a basis but it could happen that any of those polynomials be in other basis. There could also be a basis including $p_1,p_2$ and another polynyomial linealy independent to those two.If the answer was clear enough you can mark it as accepted! $\endgroup$ – Joaquin San Oct 6 '16 at 13:51

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