$Statement\;of\;the\;Problem:$ Determine the dimension of the subspace of ${\mathbf{R^4}}$ consisting of all $X\in\mathbf{R^4}$ such that: $$x_1+2x_2=0 \;\;and\;\; x_3-15x_4=0$$
$My\;Question:$ Let $V=\mathbf{R^4}$ and $W=\mathbf{R^2}$. Let $L:V\rightarrow W$ be the mapping defined by $L(X)=(x_1+2x_2,x_3-15x_4)$ where $X=(x_1,x_2,x_3,x_4)$ is in $V$. I managed to prove that the answer to the problem is 2 by proving that $L$ is surjective (onto) and therefore has a dimension of 2, which in turn gives the kernel a dimension of two though. Originally, before I could get to that proof though, I tried to see about a case by case basis for what each possible value of $dim\;Im\;L$ could be and then try to prove that it was two (the list of possible values being 0,1,2). However, when I tried to do proof by contradiction, I ran into a soft wall. $\mathbf{Issue\;1)}$ As for the case where $dim\;Im\;L=0$, I couldn't find out a way to say that this couldn't be true as that would mean the kernel of $L$ is all of $V$ which is just obviously not true by inspection. $\mathbf{Issue\;2)}$ As for $dim\;Im\;L=1$, I could never really get a proof that worked to disprove this case.
As for Issue 1, is it strong enough to just say that $dim\;Im\;L\neq0$ by inspection, or do I need to look for a more formal way to say this to be stronger? As for Issue 2, anyone got any ideas how to disprove $dim\;Im\;L=1$
[Note: If anyone wants to see I've done a proof that contains the fact that $L$ is surjective, I'll be happy to provide that as well to prove that I've already solved the problem]
Hint:
Show that the vectors $$u = (-2,1,0,0)$$ $$v = (0,0,15,1)$$ are in the solution set of the system, are linearly independent, and their span is the set of all solutions.