The vector space $R^3$ and the subset M consists of the vectors $(\xi_1,\xi_2,\xi_3)$ for which

i) $\xi_1 = 0 $

ii) $\xi_1 = 0$ or $\xi_2 = 0 $

iii) $\xi_1 + \xi_2 = 0 $

iv) $\xi_1 + \xi_2 = 1 $

Which of these spaces is itself a vector space?

I am completely stumped on how to go about figuring this out. From looking around online, I think that any $M\subset V$ is a subspace if $aX+bY$ is in $M$ for any $a,b$ scalars.

  • $\begingroup$ You must show that the zero vector is contained in the subset. The subset is closed under vector addition. The subset is closed under scalar multiplication of the vectors. $\endgroup$ – Chantry Cargill Sep 21 '14 at 15:42

A subset $M$ is a subspace if $$\forall u,v \in M:\ (\alpha\cdot u + \beta\cdot v)\in M,\ \forall \alpha,\ \beta \in \mathbb{R}$$

i) Is a subspace (Check!)

ii) Is not a subspace: $(\xi_1,0,0)\in M,\ (0,\xi_2,0)\in M, \text{ but } \\ (\xi_1,0,0)+ (0,\xi_2,0)=(\xi_1,\xi_2,0)\not\in M $

iii) Is the subspace $<(\xi_1,-\xi_1,0),(0,0,\xi_3)>$ (Check!)

iv) Is not a subspace: Since $ 0\not\in M$

  • $\begingroup$ Thanks. I am trying to understand why (ξ1,ξ2,0)∉M. Is it b/c (ξ1,0,0) and (0,ξ2,0) are zero vectors, and their addition should equal a zero vector but only one can equal zero at a time? $\endgroup$ – Snagglewhen Sep 21 '14 at 18:58

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