Originally Posted by

**Mollier** Hi,

**problem:**

Sometimes a subset of a vector space is itself a vector space (with respect to the linear operations already given). Consider, for example, the vector space $\displaystyle C^3$ and the subsets $\displaystyle V\;of\;C^3$ consisting of those vectors $\displaystyle \xi_1,\xi_2,\xi_3$ for which

(a) $\displaystyle \xi_1$ is real,

(b) $\displaystyle \xi_1=0$,

(c) either $\displaystyle \xi_1=0\;or\;\xi_2=0$,

(d) $\displaystyle \xi_1+\xi_2=0$,

(e) $\displaystyle \xi_1+\xi_2=1$

In which of these cases is $\displaystyle V$ a vector space?

**attempt:**

(a)

Not a vector space since multiplication by a scalar $\displaystyle \alpha\in C$ yields $\displaystyle \alpha x=((a+\xi_1)+bi,\alpha\xi_2,\alpha\xi_3)$.

$\displaystyle \xi_1'=(a+\xi_1)+bi$ is not in $\displaystyle V$.

(b)

Is a vector space if $\displaystyle \beta=0$. If I think of this in terms of $\displaystyle R^3$, I would have the $\displaystyle y-z$ plane if $\displaystyle \beta_1=0$ I'm not sure if I should be thinking like this..

(c)

Again, if I think of it as $\displaystyle R^3$, I would get the $\displaystyle y-z$ plane and the $\displaystyle x-z$ plane.

This is not closed under addition, $\displaystyle (0,\xi_2,\xi_3)+(\eta_1,0,\eta_3)=(\eta_1,\xi_2,\x i_3+\eta_3)$,i.e., not a vector space.

I do not see which axiom for a vector space is violated though..

(d)

Is a vector space.

(e)

Hm, I would say no since it looks like it's not closed under addition.

$\displaystyle (\xi_1,1-\xi_1,\xi_3)+(\eta_1,1-\eta_1,\eta_3)=(\xi_1+\eta_1,2-\xi_1-\eta_1,\xi_3+\eta_3)$

Sorry for my pathetic attempts, can't do much better..

Thanks.