How to find dimensions and compute bases for a cubic polynomial with constraints? I'm having problems solving this question, would appreciate any help.
The question asks to compute the dimension and find bases for the vector space - cubic polynomials $p(x, y)$, in two real variables with the properties: $p(0, 0) = 0, p(1, 0) = 0, p(0, 1) = 0$. 
Apparently you should start out with the fact that there are 10 possible dimensions for this vector space, but I don't understand why that is the case.
 A: Forget the conditions on $p$ for now - let's just think about the dimension of
$$\mathbb{P}_3(x,y):=\{p(x,y)\,:\,p(x,y)=\sum_{j+k\le3}a_{j,k}x^jy^k\}$$
A basis of $\mathbb{P}_3(x,y)$ is clearly $\mathcal{B}=\{1,x,y,x^2,xy,y^2,x^3,x^2y,xy^2,y^3\}$ so $\dim\mathbb{P}_3(x,y)=10$. If the base field is $F$, define
$$T:\mathbb{P}_3(x,y)\longrightarrow F^{10}\\
p(x,y)\mapsto(p(0,0),p(1,0),p(0,1),0,\ldots,0)$$
This is a linear transformation between vector spaces of the same dimension, so the rank-nullity theorem applies. Clearly $\operatorname{im}T$ has dimension $3$, so $\dim\ker T=7$. However $\ker T$ is precisely those polynomials in $\mathbb{P}_3(x,y)$ such that $p(0,0)=p(1,0)=p(0,1)=0$, so the space you are looking at has dimension $7$.
A basis can be found relatively easily. Suppose $p(x,y)=\sum_{j+k\le3}a_{j,k}x^jy^k\in\ker T$. Evaluating $p(0,0),p(1,0)$ and $p(0,1)$ we get
$$a_{0,0}=0,\\
a_{0,0}+a_{1,0}+a_{2,0}+a_{3,0}=0,\\
a_{0,0}+a_{0,1}+a_{0,2}+a_{0,3}=0.$$
Solving this linear system we get $a_{0,0}=0,a_{1,0}=-a_{2,0}-a_{3,0}$ and $a_{0,1}=-a_{0,2}-a_{0,3}$, all other entries as free parameters. A basis of $\ker T$ is hence
$$\{-x+x^2,-x+x^3,-y+y^2,-y+y^3,xy,x^2y,xy^3\}$$
