Confused about final answer for vector space of polynomials. a friend and I are working on some problems in linear algebra, and we have come to a bit of stump unto the correct way to look at this. The questions asks " Let T be a subspace of polynomials P2(t) spanned by, $$ p_1= 1+t+t^2 $$ 
 $$ 
p_2= 2 + 2t + 2t^2 $$
$$ p_3=-t $$
$$ p_4=1+t^2 $$(given in the question). Find a subset of the polynomials that form a basis for T.
Here is where the confusion is. I wrote the vectors as columns and row reduced, then chose the columns that had leading entries ( p1 and p3 ) ie in the form $$  \begin{matrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}$$ 
and said that these polynomials form a basis. However my friend says that in fact there are five answers, because any of the other polynomials other then {p1,p2} are linear combinations?
Is that true? To me it seems as if the only correct answer is the one with the leading entries?
Thanks all
 A: Your friend is correct.  In fact there is only one set of two vectors that you can't choose: $\{p_1, p_2\}$.
To see why multiple answers are possible try this: Write the vectors in a different order.  Specifically try switching $p_1$ and $p_4$ and then do exactly what you did before (write down a matrix, row reduce it, choose vectors corresponding to leading $1$'s).
Notice that when you write down the polynomials in the order $p_4, p_2, p_3, p_1$ your algorithm tells you to pick $\{p_4, p_2\}$.
Basically the algorithm you're using is just one way to choose vectors.  Specifically it's always going to give you the first vectors in the list that work.  When you listed them in order $p_1, p_2, p_3, p_4$ it picked $p_1$ first (it will always pick the first vector no matter what order you put these $4$ vectors in).  Then it saw it couldn't pick $p_2$ so it picked $p_3$ which was next.  When you listed them in order $p_4, p_2, p_3, p_1$ it picked $p_4$ first and picked $p_2$ second, so it chooses the first two vectors that it's allowed to choose.
