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Which of the following subsets of $\mathbb{R}^3$ constitute a subspace of $\mathbb{R}^3$?

[Here, $x = (\xi_1 \xi_2 \xi_3)$]

(a) All $x$ with $\xi_1 =\xi_2$ and $\xi_3=0$.

(b) All $x$ with $\xi_1=\xi_2+1$.

(c) All $x$ with positive $\xi_1, \xi_2, \xi_3$.

(d) All $x$ with $\xi_1-\xi_2+\xi_3=k$ const.

I have thought about the following:

For (a), denote by $M$ the space generated by these characteristics and take $(x,x,0), (y,y,0)\in M$ and $a,b \in \mathbb{R}$, with which $a(x,x,0)+b(y,y,0)=(ax+by, ax+by, 0)$ and thus $a(x,x,0)+b(y,y,0)\in M$, then $M$ is a subspace of $\mathbb{R}^3$.

For (b), denote by $M$ the space generated by these characteristics and take $(x_1,x_1-1,x_3), (y_1,y_1-1,y_3)\in M$ and $a,b \in \mathbb{R}$, with which $a(x_1,x_1-1,x_3)+b(y_1,y_1-1,y_3)=(ax_1+by_1, ax_1+by_1-(a+b), ax_3+by_3)$ and thus $a(x_1,x_1-1,x_3)+b(y_1,y_1-1,y_3)\notin M$, then $M$ is not a subspace of $\mathbb{R}^3$.

For (c), denote by $M$ the space generated by these characteristics and take $(x_1,x_2,x_3), (y_1,y_2,y_3)\in M$ and $a,b \in \mathbb{R}$ with $a<0$ and $b<0$, with which $a(x_1,x_2,x_3)+b(y_1,y_2,y_3)=(ax_1+by_1, ax_2+by_2, ax_3+by_3)$ and thus $a(x_1,x_2,x_3)+b(y_1,y_2,y_3)\notin M$, then $M$ is not a subspace of $\mathbb{R}^3$

For (d), denote by $M$ the space generated by these characteristics and take $(x_1,x_2,x_3), (y_1,y_2,y_3)\in M$ and $a,b \in \mathbb{R}$, with which $a(x_1,x_2,x_3)+b(y_1,y_2,y_3)=(ax_1+by_1, ax_2+by_2, ax_3+by_3)$ and so $ax_1+by_1-(ax_2+by_2)+ax_3+by_3=a(x_1-x_2+x_3)+b(y_1-y_2+y_3)=ac_1+bc_2=k$ const and thus $a(x_1,x_2,x_3)+b(y_1,y_2,y_3)\in M$, then $M$ is a subspace of $\mathbb{R}^3$

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  • 2
    $\begingroup$ (a) is a subspace and your proof is fine. (b) and (c) are both not subspaces, and a much shorter proof would be that they do not contain the zero vector $(0,0,0)$, but your proofs are fine (even if wordy). (d) is not a subspace unless $k=0$. For any other $k$ it will not be, again by looking at the zero vector. $\endgroup$ – JMoravitz Dec 16 '17 at 2:16
  • $\begingroup$ @JMoravitz Where does the test I did in (d) fail? $\endgroup$ – user402543 Dec 16 '17 at 2:26
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    $\begingroup$ My interpretation is that $\zeta_1-\zeta_2+\zeta_3$ must equal a very specific $k$, each and every time. E.g. if $k=2$ then we would need $\zeta_1-\zeta_2+\zeta_3=2$. You made it to $a(x_1-x_2+x_3)+b(y_1-y_2+y_3)$ which simplifies to $ak+bk=(a+b)k$ which is explicitly not equal to $k$ unless $a+b=1$ or unless $k=0$. We cannot guarantee that $a+b=1$ $\endgroup$ – JMoravitz Dec 16 '17 at 2:31
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Only (a) is a subspace, (d) is a subspace only if k=0 otherwise it doesn’t contain $\vec 0$ which is the most important one. Such argument on $\vec 0$ allow to classify not true also (b) and (c).

Indeed for a subspace all the following three properties must be satisfied:

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

thus if a subset do not contain $\vec0$, it is sufficient to conclude that it is not a subspace, this is really the most simple and first check to do.

EG

in $\mathbb{R^3}$ the only possible subset are: $\vec{0}$, the line planes throughout the origin and $\mathbb{R^3}$ itself

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