Monday, April 14, 2014

every subspace of a finite dimensional vector space is also finite dimensional

According to this:
Let $A, B$ be the bases of $W, V$ respectively. then in the vector space V, A is linearly independent, B spans V, so $\text{rank}(A) \le \text{rank}(B)$, thus completes the proof of the first part.
The second part is also straight forward. If B is a basis of V and rank(A) = rank(B), then A is also a basis of V. (review the above link.)