# There exist vectors ${v_1,…,v_m}$ such that ${u_1,…,u_k,v_1,…,v_m}$ is a basis for $V$?

I am not sure how my teacher derived this. I missed out the explanation for this.

Let ${u_1,...,u_k}$ be a basis for $V\bigcap W$.

$V$ and $W$ are both subspaces of the $R^n$

There exist vectors ${v_1,...,v_m}$ such that ${u_1,...,u_k,v_1,...,v_m}$ is a basis for $V$ and there exist vectors ${w_1,...,w_n}$ such that ${u_1,...,u_k,w_1,...,w_n}$ is a basis for $W$

-
This follows from the result that given a vector space $V$, every set of linearly independent vectors can be extended to a basis. May be he showed this in class? – Rankeya Oct 21 '12 at 3:54

First, I'm assuming we're dealing with finite dimensional spaces. Note that $V\cap W \subseteq V$ so $k \le n=\dim(V)$.

If $V=span\{u_1,\ldots, u_k\}$, then we're done. If not, take $v_1\in V-span\{u_1\,\ldots,u_k\}$ and add it to the list.

If $V=span\{u_1,\ldots, u_k,v_1\}$, we're done. If not, take $v_2\in V-span\{u_1,\ldots,u_k,v_1\}$ and add it to the list.

Continue in this fashion. Now, you should try to answer two questions:

(1.) Why does this process stop?

(2.) Is the resulting set still linearly independent?

But notice that it didn't matter too much that we used $V$ here; we can do the same process for $W$. In fact, the process we used is something we can do in general: linearly independent sets are contained in (and thus can be extended to) a basis.

-
 My grasp of your answer is still hazy at best but i will try to look at it again until i get it. Thanks anyway!!! – A New Guy Oct 21 '12 at 8:41 Let me know if you have any questions – Bey Oct 21 '12 at 14:16