# Thread: Proof related to subspaces

1. ## Proof related to subspaces

Hi guys,

I have a problem from my homework and am not sure where to start. (I hate proof questions). Could some1 just point me in the direction, I'm sure I can do it myself once I get some pointers, thanks!

2. You better get over hating proofs, because that's pretty much what math is.

Clearly each of the $\displaystyle v_i\in U_1$. If it were also in $\displaystyle U_2$, we would get $\displaystyle v_i\in U_1\cap U_2=\mathrm{Span}(u_1,\ldots,u_m)$, which would imply that the set $\displaystyle \{u_1,\ldots,u_m,v_i\}\subseteq\{u_1,\ldots,u_m,v_ 1,\ldots,v_k\}$ is not linearly independent, contradicting the fact that it is a basis for something.

Thus, for each $\displaystyle 1\leq i\leq k$, $\displaystyle v_i \notin U_2$. Since subspaces are closed under taking linear combinations of their elements, this implies $\displaystyle \{0\}=\mathrm{Span}(v_1,\ldots,v_k)\cap U_2=W\cap U_2$.

For the second part,
$\displaystyle U_1+U_2=V\Rightarrow [(U_1\cap U_2)+W]+U_2=V$
$\displaystyle \Rightarrow W+[(U_1\cap U_2)+U_2]=V$
$\displaystyle \Rightarrow W+U_2=V$.

But since we already know that the intersection of these two subspaces is trivial, this means that the sum is actually a direct sum: $\displaystyle W\oplus U_2=V$.

3. thanks alot! yeah I really should, any tips on how to improve on the thought process when tactling a proof problem?

4. Why does [U1 intersection U2)+ U2] = U2? This implies U1 intersection U2 = 0