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?

  • 3
    $\begingroup$ In your multiplication proof, what if $k=-1$? $\endgroup$ – G Tony Jacobs May 12 '17 at 18:34
  • 1
    $\begingroup$ 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. $\endgroup$ – G Tony Jacobs May 12 '17 at 18:38
  • $\begingroup$ @GTonyJacobs of course! So much work in vain. Should I delete the post? $\endgroup$ – Yos May 12 '17 at 18:39
  • $\begingroup$ 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 🤷‍♀️ $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.