Answer: The proof is done below.
Step-by-step explanation: Given that U and V are subspaces of a vector space W.
We are to prove that the intersection U ∩ V is also a subspace of W.
(a) Since U and V are subspaces of the vector space W, so we must have
0 ∈ U and 0 ∈ V.
Then, 0 ∈ U ∩ V.
That is, zero vector is in the intersection of U and V.
(b) Now, let x, y ∈ U ∩ V.
This implies that x ∈ U, x ∈ V, y ∈ U and y ∈ V.
Since U and V are subspaces of U and V, so we get
x + y ∈ U and x + y ∈ V.
This implies that x + y ∈ U ∩ V.
(c) Also, for a ∈ R (a real number), we have
ax ∈ U and ax ∈ V (since U and V are subspaces of W).
So, ax ∈ U∩ V.
Therefore, 0 ∈ U ∩ V and for x, y ∈ U ∩ V, a ∈ R, we have
x + y and ax ∈ U ∩ V.
Thus, U ∩ V is also a subspace of W.
Hence proved.