Solution:

Let's prove the given statement step by step.

Step 1: U ∩ X = {0}

First, we prove that the intersection of U and X is the zero vector space.

Assume that there exists a non-zero vector v in the intersection U ∩ X. This means that v belongs to both U and X.

Then, we have v = u1 = x1, where u1 is in U and x1 is in X.

Since U ⊆ U ⊕ X, we can write u1 as u1 = u1 + 0, where 0 is in X.

Similarly, since X ⊆ U ⊕ X, we can write x1 as x1 = 0 + x1, where 0 is in U.

Therefore, we have v = u1 = u1 + 0 = 0 + x1 = x1, which implies that v is in both U and X.

But this contradicts our assumption that v is non-zero and belongs to U ∩ X.

Hence, the intersection of U and X is the zero vector space, i.e., U ∩ X = {0}.

Step 2: U + X = U ⊕ X

Next, we prove that the sum of U and X is the direct sum U ⊕ X.

Since U ⊆ U + X and X ⊆ U + X, it is clear that U ⊕ X ⊆ U + X.

Now, let's show that U + X ⊆ U ⊕ X.

Let v be an arbitrary vector in U + X. Then, there exist vectors u in U and x in X such that v = u + x.

Since U ⊆ U ⊕ X and X ⊆ U ⊕ X, we can write u as u = u + 0, where 0 is in X, and x as x = 0 + x, where 0 is in U.

Therefore, we have v = u + x = u + 0 + x + 0 = (u + 0) + (x + 0), which implies that v is in U ⊕ X.

Hence, U + X ⊆ U ⊕ X.

Combining both inclusions, we have U ⊕ X = U + X.

Step 3: U + W = V

Finally, we prove that the sum of U and W is the entire vector space V.

Since U ⊆ V and W ⊆ V, it is clear that U + W ⊆ V.

Now, let's show that V ⊆ U + W.

Let v be an arbitrary vector in V. Since V is a finite-dimensional vector space, U ⊕ W is also a finite-dimensional vector space.

By the dimension property of finite-dimensional vector spaces, we know that dim(U ⊕ W) = dim(U) + dim(W) - dim(U ∩ W).

Since U ∩ W = {0} (from Step 1), the dimension of U ⊕ W is dim(U) + dim(W).

Since U ⊆ U ⊕ X, we have dim(U ⊕ X) ≥ dim(U).

Similarly, since W ⊆ U ⊕ X = U + X (from Step 2), we have dim(U ⊕ X) ≥ dim