Prove that if W is a subspace of a finite dimensional vector space V, then dim(W) ≤ dim(V).

To prove this, we will use the concept of a basis for a vector space. A basis for a vector space V is a set of vectors in V that are linearly independent and span V.

Step 1: Assume that W is a subspace of V. This means that every vector in W is also a vector in V.

Step 2: Since W is a subspace, it has a basis. Let's call this basis B. The vectors in B are linearly independent and they span W.

Step 3: Because every vector in W is also a vector in V (from Step 1), the vectors in B are also vectors in V.

Step 4: The vectors in B are linearly independent in V. This is because if they were not, then there would be a nontrivial linear combination of them that equals the zero vector in V. But this would contradict the fact that they are a basis for W (and hence linearly independent in W).

Step 5: The dimension of a vector space is defined as the number of vectors in a basis for that space. Therefore, the dimension of W (dim(W)) is the number of vectors in B.

Step 6: Since B is a set of linearly independent vectors in V, the number of vectors in B cannot be more than the number of vectors in a basis for V. This is because a set of more than dim(V) vectors in V must be linearly dependent.

Step 7: Therefore, the number of vectors in B (which is dim(W)) is less than or equal to the number of vectors in a basis for V (which is dim(V)).

So, we have proved that if W is a subspace of a finite dimensional vector space V, then dim(W) ≤ dim(V).