As per the title, I wish to determine which of the following subsets from $\mathbb{R}$ are vector subspaces.

I have consulted this related question, and I understand that three criteria need to be fulfilled. (namely, the 0 element is contained, the set is closed und addition and the set is closed under multiplication). Nonetheless, the notation below isn't intuitive to me. Below I have provided two subsets, as I think that the two examples are complimentary for forming an understanding.

$U_1 := \{ (x,y,z) | 3x - 4y + 2z = 0, x + 2y - z = 0\}$


$U_4 := \{ (2t + s, s, t-s) | s,t \in \mathbb{R}\}$

Any help in understanding the notation would be appreciated.


EDIT: It has been pointed out that the following proposed solution is wrong.

Proposed solution for $U_4$:

I) Subset contains the 0 element:

If $s = 0, s \in \mathbb{R} = 0.$

$s = 0 \land 2t + s = 0 \implies t \in \mathbb{R} = 0.$

II) Subset is closed under addition:

If $s = 0 \land t = 0$, then $s + t = 0$.

III) Subset is closed under multiplication:

If $s = 0 \land \lambda \in \mathbb{R}$ then $\lambda s = 0$

If $t = 0 \land \lambda \in \mathbb{R}$ then $\lambda t = 0$


What's preventing you from checking the three conditions? In the case of $U_1$:

  1. Since $3\times0-4\times0+2\times0=0+2\times0-0=0$, $(0,0,0)\in U_1$.
  2. If $3x_1-4y_1+2z_1=3x_2-4y_2+2z_2=0$, then $3(x_1+x_2)-4(y_1+y_2)+2(z_1+z_2)=0$. Also, if $x_1+2y_1-z_1=x_2+2y_2-z_2=0$, then $x_1+x_2+2(y_1+y_2)-(z_1+z_2)=0$
  3. If $3x_1-4y_1+2z_1=x_1+2y_1-z_1=0$ and $\lambda\in\mathbb R$, $3\lambda x_1-4\lambda y_1+2\lambda z_1=\lambda(3x_1-4y_1+2z_1)=0$ and $\lambda x_1+2\lambda y_1-\lambda z_1=\lambda(x_1+2y_1-z_1)=0$.

Can you do it now for $U_4$?

  • $\begingroup$ The way i understand your solution, the left hand side (LHS) of the notation contains the variables, and the right hand side represents a transformation of those variables. Is this correct? If so, for U_4 i might represent the transformation "s" as 2(s) + 1(t - s), setting both (s) and (t-s) as 0. $\endgroup$ – Oscar Mar 26 '18 at 10:11
  • $\begingroup$ @Oscar I did not think in terms of variables or of transformations. All I did was to check that: 1) $(0,0,0)\in U_1$. 2) $(x_1,y_1,z_1),(x_2,y_2,z_2)\in U_1\implies(x_1,y_1,z_1)+(x_2,y_2,z_2)\in U_1$. 3) $(x_1,y_1,z_1)\in U_1\text{ and }\lambda\in\mathbb{R}\implies\lambda(x_1,y_1,z_1)\in U_1$ $\endgroup$ – José Carlos Santos Mar 26 '18 at 10:14
  • $\begingroup$ Yes i understand. Would you mind telling me what is being represented on the left hand side and right hand side of the vertical line? $\endgroup$ – Oscar Mar 26 '18 at 10:21
  • $\begingroup$ @Oscar What are talking about? $U_1$? $U_4$? Both of them? $\endgroup$ – José Carlos Santos Mar 26 '18 at 10:23
  • $\begingroup$ Both of them. Also, I think I have shown that U_4 is indeed a vector subspace. I'll just edit my answer with my proposed solution. $\endgroup$ – Oscar Mar 26 '18 at 10:25

For a subspace all the following three properties must be satisfied:

$1) \ \vec{0} \in W\\ 2) \ \vec{v}+\vec{w} \in W\\ 3) \ \vec{cv}\to c \cdot \vec{v} \ ,c \in \mathbb{R}$

Check each one for $U_1$ and $U_4$ and you are done.

Notably for $U_1 := \{ (x,y,z) | 3x - 4y + 2z = 0, x + 2y - z = 0\}$ note that

1) $(x,y,z)=(0,0,0)\in U_1$

2) $(x_1+x_2,y_1+y_2,z_1+z_2)\in U_1$ indeed

  • $3(x_1+x_2)-4(y_1+y_2)+2(z_1+z_2)=(3x_1-4y_1+2z_1)+(3x_2-4y_2+2z_2)=0$
  • $(x_1+x_2)+2(y_1+y_2)-(z_1+z_2)=(x_1+2y_1-z_1)+(x_2+2y_2-z_2)=0$

3) $(cx+,cy,cz)\in U_1$ indeed

  • $(3cx-4cy+2cz)=c(3x-4y+2z)=0$
  • $(cx+2cy-cz)=c(x+2y-z)=0$

an for $U_4 := \{ (2t + s, s, t-s) | s,t \in \mathbb{R}\}$

1) $(s,t)=(0,0)\in U_4$

2) $(s_1+s_2,t_1+t_2)\in U_4$ indeed

  • $(2(t_1+t_2)+(s_1+s_2),s_1+s_2,(t_1+t_2)-(s_1+s_2)=(2t_1 + s_1, s_1, t_1-s_1)+(2t_2 + s_2, s_2, t_2-s_2)$

3) $(cs,ct)\in U_1$ indeed

  • $(2ct + cs, cs, ct-cs)=c(2t + s, s, t-s)$
  • $\begingroup$ The asker states in their question that they already know the criteria for being a subdpace, but don't understand the notation and execution. This answer seems to miss all that. $\endgroup$ – user296602 Mar 26 '18 at 13:13
  • $\begingroup$ @user296602 Mine was aimed to be only an hint, now I've added some detail more $\endgroup$ – gimusi Mar 26 '18 at 14:33

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