Consider the set$B=\{x^2+1,2x^2+x-1,x^2+x,2x^2+x+1\}$ I am not sure if I understand these concepts correctly.
Consider the set$B= \{x^2+1,2x^2+x-1,x^2+x,2x^2+x+1\}$ where $B$ is a subset of $P_2$.
a.) Using the concept of dimension, explain why $B$ is a linearly dependent set. - For this I just show that $\dim(B) = 4 \neq n = 3$ is this sufficient?
b.) Discard one of the vectors from $B$ to form a linearly independent set, label it $B'$. - For this I discarded the last one, so $B' = \{x^2+1,2x^2+x-1,x^2+x\}$
c.) Prove that $B'$ is a linearly independent set. - For this I just row reduced the matrix representation of $B'$, such that $B'=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ so $\dim(B')=3=n$.
d.) with no further computation explain why $B'$ is a basis for $P_2$ - The only thing I could think of here is to say that: $B'$ is a linearly independent set, and the $sp(B')=sp(P_2)$ is this correct?
I am really uncertain in all these concepts so if I'm completely wrong please let me know. Any help will be greatly appreciated! Thanks
 A: a) You are on the right track, but then again completely wrong regarding formalities: $B$ certainly does not have dimension $4$. Not only is there no four-dimensional subspace of $P_2$ at all, the set $B$ isn't even a vector space! What you meant to say was that $|B|=4$ (which is so because the four polynomials in the enumeration are pairwise distinct; an observation that is correct only if we agree that $2\ne 0$) and hence $|B|>3=\dim P_2$. As no linearly independent family can have larger cardinality than a basis, we conclude that $B$ is linearly dependent.
b) Are you aware that discarding $2x^2+x-1$ instead would have been a bad idea? You would have noticed in c)!
c) Again, $B'$ has no dimension. Either speak of the cardinality $|B'|$ of $B'$ or of the dimension $\dim\operatorname{sp}(B')$ of the space generated 
by $B'$.
d) Alternatively: In a finite-dimensional vector space, every linearly independent set of cardinality matching the dimension is already a basis. (This is because each linearly independent family can be extended to a basis, bu tI assume you have already covered that.)
