Is $L=\{(x_1,x_2,x_3,x_4) \in \mathbb{R^4} | x_1-2x_4 \ge x_2+x_3\}$ a vector space and what is its spanning set?

Is $L=\{(x_1,x_2,x_3,x_4) \in \mathbb{R^4} | x_1-2x_4 \ge x_2+x_3\}$ a vector space over $\mathbb R$ and what is its spanning set?

$\underline 0 \in L$

Let $u=(x_1,x_2,x_3,x_4)$, $v=(y_1,y_2,y_3,y_4)$; $u,v \in L$.

We have the system: $$\begin{cases} x_1-2x_4 \ge x_2+x_3\\ y_1-2y_4 \ge y_2+y_3 \end{cases}$$

If we sum up the sides we get: $$(x_1-y_1)-2(x_4+y_4) \ge (x_2+y_2)+(x_3+y_3)$$

so $L$ is closed for addition.

Regarding closure for multiplication let $k \in \mathbb R$: $$k(x_1-2x_4) \ge k(x_2+x_3) \Rightarrow kx_1-2(kx_4) \ge kx_2+kx_3$$ So $L$ is a vector subspace of $\mathbb{R^4}$.

(Not sure my proof for multiplication is good)

To find the spanning set is really tricky for me because of the inequality sign. What I did is I assumed that: $$\begin{cases} x_1-2x_4=a \\ x_2+x_3=b \\ a \ge b \end{cases}$$

Then I attempt to solve the system of equations: $$\begin{cases} x_1-2x_4=a \\ x_2+x_3=b \end{cases}$$

via the corresponding matrix (already is in reduced echelon form): $$\begin{bmatrix} 1&0&0&-2&a\\ 0&1&1&0&b \end{bmatrix}$$

So $x_3$ and $x_4$ are free coefficients, let $x_3=s, x_4=t$ then $x_1=a+2, x_2=b-s$ and we have the solution set $$\{(a+2, b-s, s,t)|a \ge b\}$$ Suppose I was correct up to here how do I derive the spanning set from the solution set?

What I did is: $$(a+2, b-s, s,t)=(a,b,s,0)+(2,-s,0,t)=$$ $$=(a,0,0,0)+(0,b,0,0)+(0,0,s,0)+(2,0,0,0)+(0,-s,0,0)+(0,0,0,t)=$$ $$=a(1,0,0,0)+b(0,1,0,0)+s(0,0,1,0)+(2,0,0,0)+s(0,-1,0,0)+t(0,0,0,1)$$

According the this: $$Sp(L)=\{(1,0,0,0), (0,1,0,0),(0,0,1,0),(2,0,0,0),(0,-1,0,0),(0,0,0,1)\}$$

But how do I enforce the $a \ge b$ condition here?

• In your multiplication proof, what if $k=-1$? – G Tony Jacobs May 12 '17 at 18:34
• To help with intuition, consider the set in $\mathbb{R}^2$ where $x_1>x_2$. It has the same problem, and the advantage that you can look at a picture of it. – G Tony Jacobs May 12 '17 at 18:38
• @GTonyJacobs of course! So much work in vain. Should I delete the post? – Yos May 12 '17 at 18:39
• I don't think that's necessary. We keep these things around for the benefit of future readers who might make the same mistake. Someone else might say otherwise 🤷‍♀️ – G Tony Jacobs May 12 '17 at 18:41

$L$ is not a vector subspace of $\Bbb R^4$ because the vector $(1,0,0,0)$ belongs to $L$, but it's opposite isn't. So $L$ is not a subgroup with respect to the inner operation, and it can not be a vector subspace.