Can this work as a basis for S?

[PeroK]

@Hall I advise posting future problems in the Homework section, where the rules and guidelines focus the helpers on helping you!

 

[fresh_42]

@Hall I advise posting future problems in the Homework section, where the rules and guidelines focus the helpers on helping you!

Sorry. I thought I would have helped by correcting the definition of $A$. I guess someone wants to scare me away.

 

[PeterDonis]

He did ***not** write $c_i = 0, i \in \{1, \dots, j\}$. Rather it was the sum of the $c_i$.

He wrote that the sum of $c_i$ for $i$ from 1 to $j$ is zero, for all values of $j$ from $1$ to $n$. That means that $c_1 = 0$ (since that's the only term in the sum for $j = 1$), and then, by induction, all of the other $c_i$ for $i$ from $1$ to $n$ must also be zero (because each time we increase $j$ by $1$ we include just one additional term, and all the previous terms are already known to be $0$ from the previous values of $j$).

That might not be what he intended to write, but that's what he wrote. If he intended to write something else, he should tell us what.

 

[PeterDonis]

a sample element of $S$ would look like:
$$
p(t) = c_0 + c_1t +c_2t^2 + \cdots + c_nt^n
$$

A sample element, yes. But for each different element of $S$, all of the $c_i$ will in general be different.

So if you are trying to find a basis for $S$, you should be thinking in terms of picking $n$ sets of values for the $c_i$, such that the resulting polynomials are all linearly independent and that any member of $S$ can be expressed as a linear combination of them.

The fact that each member of $S$ can be described by a set of $n$ numbers (the $c_i$) is what suggests the "vector" terminology. (Strictly speaking, to show that these sets of numbers actually are vectors requires proving that they satisfy the vector space axioms.)

 

[Office_Shredder]

I think fresh is right here. The op should pick the set A that they want, the set described in the original post is wrong.

Even if you pick the "right" constraint on the coefficients (where they don't all draw from the same coefficients), setting all of them equal to zero still satisfies the criteria, so no, the set A is not guaranteed to be a basis without additional constraints.