Is it a vector space?

**Mollier** · January 7th 2010, 04:59 AM

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 $C^3$ and the subsets $V\;of\;C^3$ consisting of those vectors $\xi_1,\xi_2,\xi_3$ for which
(a) $\xi_1$ is real,
(b) $\xi_1=0$ ,
(c) either $\xi_1=0\;or\;\xi_2=0$ ,
(d) $\xi_1+\xi_2=0$ ,
(e) $\xi_1+\xi_2=1$

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

attempt:
(a)
Not a vector space since multiplication by a scalar $\alpha\in C$ yields $\alpha x=((a+\xi_1)+bi,\alpha\xi_2,\alpha\xi_3)$ .
$\xi_1'=(a+\xi_1)+bi$ is not in $V$ .

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

(c)
Again, if I think of it as $R^3$ , I would get the $y-z$ plane and the $x-z$ plane.
This is not closed under addition, $(0,\xi_2,\xi_3)+(\eta_1,0,\eta_3)=(\eta_1,\xi_2,\x i_3+\eta_3)$ ,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.
$(\xi_1,1-\xi_1,\xi_3)+(\eta_1,1-\eta_1,\eta_3)=(\xi_1+\eta_1,2-\xi_1-\eta_1,\xi_3+\eta_3)$

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

Thanks.

**tonio** · January 7th 2010, 05:54 AM

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 $C^3$ and the subsets $V\;of\;C^3$ consisting of those vectors $\xi_1,\xi_2,\xi_3$ for which
(a) $\xi_1$ is real,
(b) $\xi_1=0$ ,
(c) either $\xi_1=0\;or\;\xi_2=0$ ,
(d) $\xi_1+\xi_2=0$ ,
(e) $\xi_1+\xi_2=1$

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

attempt:
(a)
Not a vector space since multiplication by a scalar $\alpha\in C$ yields $\alpha x=((a+\xi_1)+bi,\alpha\xi_2,\alpha\xi_3)$ .
$\xi_1'=(a+\xi_1)+bi$ is not in $V$ .

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

(c)
Again, if I think of it as $R^3$ , I would get the $y-z$ plane and the $x-z$ plane.
This is not closed under addition, $(0,\xi_2,\xi_3)+(\eta_1,0,\eta_3)=(\eta_1,\xi_2,\x i_3+\eta_3)$ ,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.
$(\xi_1,1-\xi_1,\xi_3)+(\eta_1,1-\eta_1,\eta_3)=(\xi_1+\eta_1,2-\xi_1-\eta_1,\xi_3+\eta_3)$

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

Thanks.

Questions like this don't make much sense unless you specify over what field is the parent vector space defined. Usually, when talking of $\mathbb{C}^3$ we take this as a complex v.s., i.e. defined over $\mathbb{C}$ ...but you need to specify this.

Tonio

**HallsofIvy** · January 7th 2010, 06:22 AM

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 $C^3$

I assume that $C^3$ is the vector space of ordered triples of complex numbers with the usual addition and scalar multiplication.

[quote] and the subsets $V\;of\;C^3$ consisting of those vectors $\xi_1,\xi_2,\xi_3$ for which
(a) $\xi_1$ is real,
(b) $\xi_1=0$ ,
(c) either $\xi_1=0\;or\;\xi_2=0$ ,
(d) $\xi_1+\xi_2=0$ ,
(e) $\xi_1+\xi_2=1$

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

attempt:
(a)
Not a vector space since multiplication by a scalar $\alpha\in C$ yields $\alpha x=((a+\xi_1)+bi,\alpha\xi_2,\alpha\xi_3)$ .
$\xi_1'=(a+\xi_1)+bi$ is not in $V$ .[quote]
Are you assuming that $\alpha= a+ bi$ ? If so, you should say so. And you mean that $a\xi_1+ bi$ is not real, not that it is not in V. V is a set of vectors in $C^3$ and $a\xi_1+ bi$ is a single number. A simpler example is to take $\alpha= i$ .

[quote]
(b)
Is a vector space if $\beta=0$ . If I think of this in terms of $R^3$ , I would have the $y-z$ plane if $\beta_1=0$ I'm not sure if I should be thinking like this..[/tex]
If $\xi_1= 0$ you are left with vectors of the form $(0, \xi_2, \xi_3)$ . What do you get if you add two such things? Multiply by some complex number?

(c)
Again, if I think of it as $R^3$ , I would get the $y-z$ plane and the $x-z$ plane.
This is not closed under addition, $(0,\xi_2,\xi_3)+(\eta_1,0,\eta_3)=(\eta_1,\xi_2,\x i_3+\eta_3)$ ,i.e., not a vector space.
I do not see which axiom for a vector space is violated though..

Both (0, 1, 1) and (1, 0, 1) are in this subset. Is their sum?

(d)
Is a vector space.

Yes, that is correct. I assume you have shown that.

(e)
Hm, I would say no since it looks like it's not closed under addition.
$(\xi_1,1-\xi_1,\xi_3)+(\eta_1,1-\eta_1,\eta_3)=(\xi_1+\eta_1,2-\xi_1-\eta_1,\xi_3+\eta_3)$

Again, both (1, 0, 1) and (0, 1, 0) are in this subset. Is their sum?

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

Thanks.

**Mollier** · January 7th 2010, 08:11 PM

Originally Posted by tonio

Questions like this don't make much sense unless you specify over what field is the parent vector space defined. Usually, when talking of $\mathbb{C}^3$ we take this as a complex v.s., i.e. defined over $\mathbb{C}$ ...but you need to specify this.
Tonio

The question was taken from Finite-Dimensional Vector Spaces by Halmos. I guess he assumes the same thing as HallsofIvy..
Thanks for the heads up!