If U and W are subspaces of a vector space V, show that and are subspaces.

Answer 1

To show that the sum U + W is a subspace of V, we need to verify three things:

1) U + W is non-empty.
2) U + W is closed under addition.
3) U + W is closed under scalar multiplication.

1) Since U and W are both subspaces of V, they are non-empty. Therefore, there exists at least one element in both U and W. Let's call these elements u and w, respectively. Since u ∈ U and w ∈ W, we have u + w ∈ U + W. Thus, U + W is non-empty.

2) To show that U + W is closed under addition, we need to show that for any u₁, u₂ ∈ U and w₁, w₂ ∈ W, their sum u₁ + w₁ + u₂ + w₂ ∈ U + W.

Since U is a subspace, it is closed under addition. Therefore, u₁ + u₂ ∈ U. Similarly, since W is a subspace, w₁ + w₂ ∈ W.

Since u₁ + u₂ ∈ U and w₁ + w₂ ∈ W, their sum (u₁ + u₂) + (w₁ + w₂) ∈ U + W.

Therefore, U + W is closed under addition.

3) To show that U + W is closed under scalar multiplication, we need to show that for any scalar k and vector v ∈ U + W, their product kv ∈ U + W.

Let v be a vector in U + W. This means that there exist u ∈ U and w ∈ W such that v = u + w.

Since U is a subspace, it is closed under scalar multiplication. Therefore, ku ∈ U.

Similarly, since W is a subspace, kw ∈ W.

Since ku ∈ U and kw ∈ W, their sum ku + kw = k(u + w) = kv ∈ U + W.

Therefore, U + W is closed under scalar multiplication.

Since U + W satisfies all three properties of being a subspace, we can conclude that U + W is a subspace of V.