show that the only subspaces of R1 are 0 and R1 itself.
okay , so I don't know if I'm making it more complicated than it is, but R1 is just a point right? so i don't know how i would prove this?
To prove that the only subspaces of R1 are the zero vector space {0} and R1 itself, we need to consider the definition and properties of subspaces.
A subspace of a vector space V is a subset of V that is closed under vector addition and scalar multiplication, and it also contains the zero vector. In the case of R1, it is a one-dimensional vector space consisting of real numbers.
1. First, let's consider the zero vector space {0}. This set contains only the zero vector, which is the additive identity element of R1. Since it contains the zero vector, it automatically satisfies the requirement for a subspace.
2. Next, let's consider R1 itself, which is the entire one-dimensional vector space. Any vector in R1 can be written as c * 1, where c is a real number and 1 is the only basis vector in R1. Since R1 includes all possible scalar multiples of the basis vector 1, it is closed under scalar multiplication. Additionally, R1 is also closed under vector addition because the sum of any two real numbers is still a real number.
Now, let's prove that these are the only subspaces of R1:
Assume there exists a subspace W of R1 that is not {0} or R1 itself. Since W is a subspace, it must contain the zero vector.
1. If W only contains the zero vector, then W = {0}, which is the zero vector space.
2. If W contains a nonzero vector v, then it must also contain all scalar multiples of v, due to the closure under scalar multiplication property. Therefore, W must contain all possible vectors in R1, which means W = R1 itself.
In either case, we conclude that the only subspaces of R1 are {0} and R1 itself.