Let $V_n$ be the vector space consisting of all polynomials of the form $$f(x,y)=\sum_{i=0}^n\sum_{j=0}^n a_{i,j}x^iy^j$$ where $a_{i,j}\in\mathbb{R}$.
(a) State the dimension of $V$, and give a basis for $V$.
(b) Let $U\leq V$ be the subspace of $V$ consisting of all $f\in V$ such that $f(x,x)=0$ for all $x\in\mathbb{R}$. Compute a basis for $U$, and determine the dimension of $U$.
Part (a) is relatively straightforward. The dimension is $(n+1)^2$. I am looking for assistance in how to express the basis. It is clear that a basis is composed of all monic polynomials of order $0$ to $n$ in $x$ and $y$. So, something along the lines of $\{1,x^i,y^j,x^iy^j:i,j=1,2,\cdots,n\}$, maybe?
For part (b), without the $a_{i,j}$ coefficients, $U$ would just be the trivial subspace, but I'm not sure otherwise.
You're correct about $(a)$.
For part $(b)$, note that $U$ is the kernel of the linear map $$V\to \Bbb R^{2n+1}:\sum_{i=0}^n\sum_{j=0}^n a_{i,j}x^iy^j\mapsto \left(\sum_{i+j=N} a_{i,j}\right)_{N=0\ldots 2n+1}$$ This map is surjective, which gives you the dimension of $U$. For a basis, fix the value of $N$, set $a_{i,j}=0$ when $i+j\neq N$ and you are left essentially to compute a basis of the kernel of $$\Bbb R^{N+1}\to \Bbb R:(x_0,\ldots ,x_{N})\mapsto \sum_{i=0}^N x_i$$