Intersection of real polynomial subspaces of degree at most 4. Let P4 be the vector space of real polynomials of degree at most 4.
$U := \{p ∈ P4 : p(−1) = p(1) = 0\}$
$V := \{p ∈ P4 : p(1) = p(2) = p(3) = 0\}$
1) Determine the subspace $U ∩ V$ .
2) Describe bases for $U, V$ and $U ∩ V$
also how do you know if the sets $U$ & $V$ are subspaces?
 A: The subspace $U \cap V$ is given by "and-ing" the conditions on $U$ and $V$. In other words, $$U \cap V = \{ p \in P_4 \;|\; p(-1)=p(1)=0 \mbox{ and } p(1)=p(2)=p(3)=0 \}$$
So anything in $U \cap V$ is a fourth degree (or less) polynomial with roots $x=-1,1,2,3$. This means that if $f(x) \in U \cap V$, then $f(x)$ must have factors $(x+1)$, $(x-1)$, $(x-2)$, and $(x-3)$. The only such polynomials (of degree four or less) are $f(x)=c(x+1)(x-1)(x-2)(x-3)$. So $U \cap V$ is multiples of $(x+1)(x-1)(x-2)(x-3)$. This polynomial would also be a good choice for a basis for $U \cap V$.
As for bases for $U$ and $V$, I'll discuss $U$ and leave $V$ to you. The same reasoning applied to $U \cap V$ applies to $U$. Any $f(x) \in U$ must have roots $x=-1$ and $1$. In other words, it must be divisible by $(x+1)(x-1)=x^2-1$. Since $U$ consists of polynomials of degree 4 and less, take $x^2-1$ and hit it with a basis for polynomials up to degree $4-2=2$. This is $\{1,x,x^2\}$ so that $\{x^2-1,x^3-x,x^4-x^2\}$ is a basis for $U$. This means that $U$ consists of linear combinations of these guys, so its elements look like $(ax^2+bx+c)(x^2-1)$ (i.e. polynomials of degree 4 or less with roots at least at $x=-1$ and $x=1$).
Finally, $U \cap V$ is a subspace because intersections of subspaces are always subspaces. $U$ and $V$ are subspaces themselves because they are solution sets for systems of homogeneous equations (i.e. kernels or nullspaces).
