What does dimension of polynomial mean? So, I know that the vector space of polynomials with degree $n$ has dimension $n+1$. What does this exactly mean? 
I'm asking specifically because of the following question (from Putnam and Beyond): 
Find the determinant of:
$$
\left[\begin{array}{ccc}
1^k & 2^k & 3^k & \dots & n^k \\
2^k & 3^k & 4^k & \dots & (n+1)^k \\
\vdots & \vdots & \vdots & \ddots & \vdots \\ 
n^k & (n+1)^k & (n+2)^k & \dots & (2n-1)^k \end{array}\right].
$$
The solution said:
"The polynomials $P_j(x) = (x+j)^k, j = 0, 1,\ldots,n−1,$ lie in the $(k+1)$-dimensional real vector space of polynomials of degree at most $k$. Because $k+1 < n$, they are linearly dependent. The columns consist of the evaluations of these polynomials at $1, 2,\ldots,n$, so the columns are linearly dependent. It follows that the determinant is zero."
I don't really understand how we can go from the polynomials to the evaluations of the polynomials. Specifically, wouldn't the evaluations be specifically in $\mathbb{R}$ and thus be in a one-dimensional space?
 A: I just got the answer - we don't have to evaluate it since we know that 
\begin{array}{ccc}
x^k & (x+1)^k & (x+2)^k & \dots & (x+n-1)^k \\
x^k & (x+1)^k & (x+2)^k & \dots & (x+n-1)^k \\
x^k & (x+1)^k & (x+2)^k & \dots & (x+n-1)^k \\
x^k & (x+1)^k & (x+2)^k & \dots & (x+n-1)^k \\
\end{array}
is linearly dependent, so any evaluation is linearly dependent.
A: It means that (irrespective of the particular field used) the vector space of one-variable polynomials of degree not more than $n$ has a basis of $n+1$ polynomials (linearly independent and spanning).
Suppose $F$ is our field and $x$ is our one variable.  Then the vector space is:
$$ V = \{ a_0 + a_1 x + \ldots + a_n x^n \;|\; a_0,a_1,\ldots,a_n \in F \} $$
with the usual addition of polynomials and usual multiplication of coefficients being the respective vector space operations.
Clearly $\{1,x,\ldots,x^n\}$ is a spanning set (any polynomial of degree at most $n$ is a linear combination of these monomial terms).  It remains to prove linear independence.
If we are thinking of polynomials formally, as algebraic constructs, then linear independence is given to us by construction.  Two polynomials are equal (in the algebraic sense) iff their coefficients are equal, because the polynomial is defined by its coefficients.
A trickier situation exists if you take the polynomials not as their algebraic expressions, but as functions.  In that case you need to be more explicit about the field $F$ or whatever is the domain assumed (commonly) for these polynomial functions.  However if $F$ is an infinite field, or if the domain used by these functions is an infinite field extension of $F$, then it can also be proven that the above set of monomials is linearly independent.
Thus in the algebraic setting we have a basis of $n+1$, and the proof can also be made in many useful cases where the polynomials are functions rather than formal algebraic constructs.
