0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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$

$\endgroup$
  • $\begingroup$ 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

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.