How to find a subset that contains all linearly independent polynomials?

Question

I found that a set S is linearly independent. How can I find a subset A of S that contains all linearly independent polynomials?

My set S consists of the following polynomial vectors in P3:

pv1 = <1, -2, -1, 0> pv2 = <0, 1, 1, 0> pv3 = <1, -1, 0, 1> pv4 = <3, 4, 1, 4>

where the first number is for a polynomial coefficient, and the rest are values for t, t squared, and t cubed for a variable t in the polynomial.

Answer 1

Proved that $S$ is linearly independent set of vectors, any subset $A\subseteq S$ works.

You can prove the next proposition:

Let $V$ a vector space, and let $S$ a linearly independent subset of $V$. Then for any $v\in S$ we have that $S-\{v\}$ is linearly independent.

One can prove this easily by contradiction. Suppose that $S':=S-\{v\}$ is linearly dependent. Then there exists $a_1,\dotsc,a_n$ not all equal to zero such that $a_1v_1+\dotsc+a_nv_n=0$ for any vectors $v_1,\dotsc,v_n$ in $S'$. Thus the scalars $a_1,\dotsc,a_n,0$ are a non-trivial solution for $a_1v_1+\dotsc+a_nv_n+0v=0$; so $S$ is linearly dependent, a contradiction.