# Subsets U of the vector space V

• AkilMAI
In summary, to find the base and dimension of U in V = P3, we can use the fact that p(x) must satisfy the condition p'(0) = p(1), which leads to a set of equations for the coefficients of p(x). By solving these equations, we can determine the base and dimension of U.

## Homework Statement

How can I find the base and dim of U here?, V = P3; U = {p in P3 : p'(0) = p(1)}...

## The Attempt at a Solution

now I've proven it is a subspace and that it is closed under addition and scalar multiplication...but how can I find the base and dim?I was thinking about writing it as p(x)=a+bx+cx^2+dx^4=>p(x)'=b+2cx+3dx^2
and...p'(0)=p(1)=>b=a+b+c+d=>d=-a-c...
Thank you

AkilMAI said:

## Homework Statement

How can I find the base and dim of U here?, V = P3; U = {p in P3 : p'(0) = p(1)}...

## The Attempt at a Solution

now I've proven it is a subspace and that it is closed under addition and scalar multiplication...but how can I find the base and dim?I was thinking about writing it as p(x)=a+bx+cx^2+dx^4=>p(x)'=b+2cx+3dx^2
and...p'(0)=p(1)=>b=a+b+c+d=>d=-a-c...
Thank you

Let p(x) be a function in P3, which means that p(x) = a + bx + cx2 + dx3.

For p(x) to be in U, it must be true that p'(0) = p(1), which implies that, and as you found, b = a + b + c + d ==> a + c + d = 0.

Solve for a to get
a = -c - d
b = b
c = c
d = d

The last three equations are trivially true.
From this set of four equations, what can you say about the coefficients of a function p(x) that belongs to U?