Prove that odd polynomials $f(x)$ of degree $\leq 10$ with $f(-1) = 0$ form a vector space. 
Let $P(X)$ be the usual vector space of polynomials in $x$ with real coefficients.
Let $U$ denote the subset of $P(X)$ consisting of those elements $f(x)$ which have degree less than or equal to $10$, satisfy $f(-x)=-f(x)$ for all $x$ and also satisfy $f(-1)=0$.
Prove $U$ is a subspace of $P(x)$.
Find a basis for $U$ and the dimension of $U$.

I have proved that $U$ is a subspace of $P(X)$ but am not sure how to find a basis and dimension of $U$

$f(-x)=-f(x)$ means that the subspace $U$ consists solely of odd functions. The subspace $V$ of odd functions in $P(X)$ with degree less than or equal to 10 has basis $x,x^{3},x^{5},x^{7},x^{9}$. Indeed, all these functions are linearly independent, odd and have degree less than 10. Moreover, they span $V$: If $g \in P(X)$ has degree less than or equal to $10$, then $g(x)=a_{10}x^{10}+a_{9}x^{9}+...+a_{1}x+a_{0}$ for some $a_{i}\in \mathbb{R}, 0 \leq i \leq10$. Now $g(-x)=-g(x)$ says $a_{i}=-a_{i}$ if $i$ is even, so these must be $0$. Then $g$ is an $\mathbb{R}$-linear combination of $x,x^{3},x^{5},x^{7},x^{9}$. So $dim(V)=5$.
Notice that $x^3-x, x^5-x, x^7-x, x^9-x$ all lie in $V$, are linearly independent and are all attain the value $0$ at $x=1$ (hence at $-1$ as they are all odd functions). So $dim(U)\geq 4$. But the function $x$ is odd and doesn't equal $0$ at $x=1$, so $U$ can't be all of $V$. So $dim(U)=4$, with a basis given by $x^3-x, x^5-x, x^7-x, x^9-x$. $\square$
