Proof related to subspaces

**holaboo** · February 12th 2011, 04:09 PM

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!

**topspin1617** · February 12th 2011, 04:43 PM

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

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

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

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

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

**holaboo** · February 13th 2011, 05:24 AM

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

**holaboo** · February 13th 2011, 08:55 AM

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

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