# Thread: Subspace Test Proof (of test not subset)

1. ## Subspace Test Proof (of test not subset)

On my midterm review there was a question asking me to:
State and prove the subspace test.

Now, I know how the subspace test works and how to use it to test whether subsets are subspaces or not...but how would I go about proving that the subspace test works?

Would the reasoning for the subspace test be that the only 3 vector space properties that need to be checked in order to declare a subset S of a vector space V a subspace of V (a subset of V that is a vector space with the same operations as V) are "x + y is defined and in S", "kx is defined and in S", and "there is a zero element"?

2. Originally Posted by crymorenoobs
On my midterm review there was a question asking me to:
State and prove the subspace test.

Now, I know how the subspace test works and how to use it to test whether subsets are subspaces or not...but how would I go about proving that the subspace test works?

Would the reasoning for the subspace test be that the only 3 vector space properties that need to be checked in order to declare a subset S of a vector space V a subspace of V (a subset of V that is a vector space with the same operations as V) are "x + y is defined and in S", "kx is defined and in S", and "there is a zero element"?
What is the subspace test? If I am understanding you a lot of properties for any subset of vector space are inherited from the vector space itself (i.e. commutativity, distributivity, assoctiativity...)

3. Sorry, I probably should have clarified that..

A subset S of a vector space V is a subspace of V if:
- S contains a zero element
- for any x in S and y in S, x + y is in S
- for any x in S and scalar k, kx is in S

I think that these are the only 3 vector space properties affected by taking a subset of the vector space...but I am not sure about that.

It would make sense since the addition and scalar multiplication properties are the only 2 that ensure the result is still within the vector space (in this case, the new subspace), and the 0 element would need to be checked since a subset of the vector space could exclude the 0 element in some cases.