Q:

Dimension of a linear space of polynomials

Let $P$ be a linear space of polynomials $P(x,y)$ of degree $\le2013$. Its subspace $V$ is formed by those polynomials for which the line integral $\oint_{x^2+y^2=R^2}f(x,y)ds=0$ for all R. What is the dimension of $V$?

A:

If you mean that $P=\operatorname{span}\{ x^i y^j \}_{i+j \le 2013}$ then $\dim P = \binom{2015}{2}$.
In selecting $f(x,y) \in P$ so that $\oint_{x^2+y^2=R^2}f(x,y)ds=0$, the coefficients of $x^i y^j$ for $0 < i+j \le 2013$ may all be independently chosen, but, once chosen, they determine a unique choice for the constant term. Thus the dimension of $V$ is $\binom{2015}{2} - 1$.
The line integral is a linear functional, i.e. you could express it as a $1 \times \binom{2015}{2}$ matrix that you multiply on the right by a vector of coefficients of the polynomial to be integrated. Since it is not a zero matrix, its rank is 1 and the dimension of its null space is $\binom{2015}{2} - 1$.

