SOLUTION: I have a Linear Algebra question. If Q is the vector space where f(x)=f^2(x) and f(x)=1 is Q a real vector space? Assuming it is a subspace of F(Negative Infinity to Infinity) w

Algebra ->  College  -> Linear Algebra -> SOLUTION: I have a Linear Algebra question. If Q is the vector space where f(x)=f^2(x) and f(x)=1 is Q a real vector space? Assuming it is a subspace of F(Negative Infinity to Infinity) w      Log On


   



Question 867004: I have a Linear Algebra question.
If Q is the vector space where f(x)=f^2(x) and f(x)=1 is Q a real vector space?
Assuming it is a subspace of F(Negative Infinity to Infinity) we only need to check the 2 axioms addition and scalar multiplication
1. Addition f(x) + f(x) = 1 + 1 = 2
Here is my question: f(x) = 2 isn't within the space, does that mean we don't have a Real vector space?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct.


According to this link, it says

"Closure: If u and v are any vectors in V, then the sum u + v belongs to V."

Both f(x) = 1 and g(x) = 1 belong to the vector space Q, but h(x) = f(x) + g(x) = 2 does NOT belong to the vector space Q (since h^2(x) = 4, h(x) = 2, h^2(x) doesn't equal h(x))

So because the closure rule doesn't hold for all elements in Q, this means Q is NOT a vector space.