- #1

Essnov

- 21

- 0

Prove: Any subspace of a finite-dimensional vector space is finite-dimensional.

Here's my attempt. I am not sure about some details and I'm hoping you guys can tell me if it makes sense or suggest improvements, give some feedback, suggest a better method, etc.

Let V be a finite-dimensional vector space such that dim(V) = n.

Now let W be a subspace of V and let S be a linearly independent set of vectors {w

_{1}, w

_{2}, ... , w

_{r}} in W.

If S spans W, then S is a basis for W and W has dimension r. If S does not span W, then add a new vector from W to S such that S is still linearly independent. If W is infinite-dimensional, then we should be able to add as many vectors as we like and S will still not span W.

However, since all vectors in W are also in V, then S is also a set of vectors in V. Therefore, S can contain at most n vectors while still being linearly independent. And if S contains n vectors, then S spans V and thus W as well.

This implies that W has dimension of at most n, and thus is finite-dimensional.

Thanks in advance for any comments.