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$Statement\;of\;the\;Problem:$ 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]

Johnq
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  • A proof by contradiction is not the best choice. Instead, just find a basis (i.e., find two vectors in $\mathbb{R}^4$ which are linearly independent and span the space of solutions to the given system of equations). Your other proof strategy (using image and kernel) is fine. – quasi Jun 23 '17 at 18:34
  • @quasi : Ok, not something I considered before (pause for me to bang head on wall in shame...). But to ask, that would assume that you know that it's only two vectors that you need to look for in the first place. How do you prove that it's only two vectors that you need to look for? Playing Devil's Advocate and forgetting obvious intuitive factors that lead to 2 as an obvious first choice (e.g. the fact that the coordinates can be solved in terms of each other to give two independent degrees of freedom), what would be the argument that says that one vector couldn't generate the subspace? – Johnq Jun 23 '17 at 18:47
  • The variables $x_2$ and $x_4$ are free. You can choose them arbitrarily. Once chosen, $x_1$ and $x_3$ are determined. Exactly two degrees of freedom. – quasi Jun 23 '17 at 18:50
  • And then find a basis that makes that intuition rigorous. – quasi Jun 23 '17 at 18:53

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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.

quasi
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  • Ok. I think I get it now. Even if you were to assume 1 vector did span the space, that can always be decomposed into two linearly independent vectors, each dictated by one coordinate. So you still get a dimension of two. Thank you very much. – Johnq Jun 23 '17 at 19:05
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    It's more like this: If you assume one nonzero vector, $w$ say, generates the space, then both $u$ and $v$ would have to be in $\langle{w}\rangle$ (the span of ${w}$). But that means $u,v$ would be parallel to $w$, hence parallel to each other, which is clearly not the case. Alternatively, given that you've verified that $u,v$ are linearly independent, the invariance of dimension principle guarantees that the span of the set ${u,v}$ can't be generated by one element. – quasi Jun 23 '17 at 19:18
  • I like that too. :) Elegant. – Johnq Jun 23 '17 at 19:20