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Let $U$ := {($x_1,x_2,x_3,x_4)|x_1 - 2x_2+ x_3 - x_4 = 0$} be a linear subspace of the space $\mathbb{R}^4$ and $v,w \in \mathbb{R}^4$ arbitrary.

Can $U$ be written in the form Span{v,w}?

While intuitevly the answer is pretty easy (It can not be is the answer), I struggle a little bit with a formal proof of this fact.

What i said is basically that U is the Solution set of this linear equation and since this equation is a "system of equations" in normal form, 3 variables are freely choosable.
Therefore, we can do the following:
$x_2 := b$
$x_3 := c$
$x_4 := d$
$=> x_1 = 2b-c+d$
So U can be written as: {$2b-c+d,b,c,d | b,c,d \in \mathbb{R}^4$}
U = {$b \times (2,1,0,0) + c \times (-1,0,1,0) + d \times (1,0,0,1) |b,c,d \in \mathbb{R}^4$}
= $\mathbb{R} \times (2,1,0,0) + \mathbb{R} \times (-1,0,1,0) + \mathbb{R} \times (1,0,0,1)$

Since these three vectors are linear independent, they are the basis of a vector subspace. No smaller basis can be found because different basis always have the same amount of vectors. So Span(v,w) does not exist because $v,w$ would have to be a basis of $U$.

Excuse my proof-writing skills. I am not really statisfied with it, I dont even know if this is correct. What can I do better?

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  • $\begingroup$ In this case, you don't need a proof. You need a counter-example to the claim "$U$ can be written as $\text{span}\{u,v\}$ for any two vectors $u,v\in\Bbb R^4$". It's easy to get such a counter-example. Just consider a vector $w\in\Bbb R^4$ such that $w\notin U$ and you can use $\text{span}\{w,2w\}$ as your counter-example. $\endgroup$ – learner Dec 13 '15 at 1:47
  • $\begingroup$ Excuse me, I did not make this very clear. I am not a native speaker. By "Let v,w be arbitary" I meant you can choose them to your liking to fit the condition. $\endgroup$ – LionIsLoose Dec 13 '15 at 9:22
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For every $x=(a,b,c,d)\in R^4$ let $f(x)=a-2b+c-d.$ We have $x\in U\iff f(x)=0. $ And $U\ne R^4$. For example $(1,0,0,0)\not \in U.$Choose any $y\in R^4\backslash U. $ Note that $f(y)\ne 0.$ $$\text { Let } S=\{yr+u : r\in R\wedge u\in U\}.$$ $$\text {Claim : } S=R^4.$$ Proof : For any $x\in R^4 $ let $r_x=f(x)/f(y)$. Let $u_x=x-y r_x$. We have $u_x\in U $ because $f(u_x)=0,$ and we have $$x=yr_x+u_x\in S.$$ Now if there were $y',y''\in R^4$ such that $\{y's+y''t: s,t\in R\}=U,$ $$\text {then } \{y r+y's+y''t :r,s,t\in R\}=S=R^4$$ which implies that $R^4$ has a vector-space basis with at most 3 members.

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