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I'm working through the following problem.

Let $U = \{ p \in \mathbb{P}_3 : p(1) = 0 \}$ and $V = \{ p \in \mathbb{P}_3 : p(-1) = 0 \}$. Here, $\mathbb{P}_3$ represents the space of polynomials of at most degree 3.

What are the dimension of $U$ and $V$ respectively?

Determine a basis for the subspace $U \cap V$.

Determine $U + V$.

How does one approach this problem? I chose the basis for $\mathbb{P}_3$, $\beta = \{x^3, x^2, x, 1 \}$. My instinct is $dim(U) = 3$ and $dim(V) = 3$ because looking at $\beta$, the constant polynomial basis vector will never have a non-zero coefficient on its own in $U$ or $V$. But how do I show this?

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Any polynomial in $\mathbb{P}_3$ is of the form $ax^3+bx^2+cx+d$ where $a,b,c,d\in\mathbb{R}$. If $p(1)=0$, then $a+b+c+d=0$, so we can write $d=-a-b-c$, and hence polynomials in $U$ are of the form $ax^3+bx^2+cx-a-b-c=a(x^3-1)+b(x^2-1)+c(x-1)$. The dimension of $U$ is therefore $3$, since $\{x^3-1,x^2-1,x-1\}$ is a basis for $U$. Showing $\dim V=3$ is similar.

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  • $\begingroup$ Excellent answer, thank you. Once I have bases for $U$ and $V$, how do I go about finding a basis for $U \cap V$? $\endgroup$ – workwork Dec 17 '17 at 20:01
  • $\begingroup$ If $ax^3+bx+cx+d\in U\cap V$, then $d=-a-b-c$ (condition from $U$) and $d=a-b+c$ (condition from $V$), so $a-b+c=-a-b-c$, which implies $c=-a$. This means every polynomial in $U\cap V$ is of the form $ax^3+bx^2-ax-b$, so the basis is $\{x^3-x,x^2-1\}$. $\endgroup$ – A. Goodier Dec 17 '17 at 20:11

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