prove that if U and W are vector subspaces of a vector V, then UV, is a vector subspace of V

To prove that the sum of two vector subspaces UV is also a vector subspace of V, we need to show the three properties of a vector subspace:

1. Closure under addition: For any two vectors u and v in UV, their sum u + v must also be in UV.
To prove this, let u and v be arbitrary vectors in UV. Since u and v are in UV, they can be written as linear combinations of vectors in U and W respectively.
u = u1 + u2 + ... + un, where u1, u2, ..., un are vectors in U.
v = w1 + w2 + ... + wn, where w1, w2, ..., wn are vectors in W.
Now, consider the sum u + v. By substituting the above expressions for u and v, we have:
u + v = (u1 + u2 + ... + un) + (w1 + w2 + ... + wn)
= (u1 + w1) + (u2 + w2) + ... + (un + wn)
Since U and W are vector subspaces, u1 + w1, u2 + w2, ..., un + wn are also in U and W respectively.
Hence, u + v can be expressed as a linear combination of vectors in U and W, which means u + v is in UV.

2. Closure under scalar multiplication: For any vector u in UV, and any scalar c, the scalar multiple c * u must also be in UV.
To prove this, let u be an arbitrary vector in UV. Then, u can be written as a linear combination of vectors in U and W:
u = u1 + u2 + ... + un, where u1, u2, ..., un are vectors in U.
Now, consider the scalar multiple c * u. By distributing the scalar c, we have:
c * u = c * (u1 + u2 + ... + un)
= c * u1 + c * u2 + ... + c * un
Since U is closed under scalar multiplication, c * u1, c * u2, ..., c * un are also in U.
Hence, c * u can be expressed as a linear combination of vectors in U and W, which means c * u is in UV.

3. Contains the zero vector: The zero vector 0 must be in UV.
Since U and W are both vector subspaces, they both contain the zero vector. Therefore, the zero vector is common to both U and W. Hence, it is also in UV.

By satisfying all three properties of a vector subspace, we have proven that UV is a vector subspace of V when U and W are vector subspaces of V.