I have a problem that says...
Show that $span((4,0),(0,-1))=R^2$
I have been working on this problem and this is what I have so far...
Let m $ =(4,0)$ and n $ =(0,-1) $then
$c1(4,0) + c2(0,-1) = (x,y)$
$(4c1, 0c1) + (0c2, -c2) = (x,y)$
$4c1+0c2=x ----> c1=x/4$
$0c1-c2=y ----> c2=-y$
I understand that a span is the set of all linear combinations for vectors, but I'm not sure if I'm proving this right? I feel like I've shown the span is in $R^2 $ but I also feel like my proof is incomplete. Is this how you do this problem? If not, could you please explain.
Your proof is basically correct. You have shows that you can write any vector $\pmatrix{x \\ y}$ as a linear combination of the two vectors given: $$ \pmatrix{x \\ y} = \frac{x}{4}\pmatrix{4 \\ 0} + (-y)\pmatrix{0\\ -1}. $$ Do indeed the two vectors span all of $\mathbb{R}^2$. If you are writing this down (in for example a homework asignment), then you might start the proof by saying "Let $\pmatrix{x \\ y} \in \mathbb{R}^2$ be given."
Another way you can make the arguments (this depends on what you already "know") is to say/show that the two vectors are linearly independent. Two linearly independent vectors in $\mathbb{R}^2$ will span all of $\mathbb{R}^2$.