Thread: Is it a vector space?

1. Is it a vector space?

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.

2. 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

3. 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.

4. 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..

Originally Posted by HallsofIvy
(a)
I assume that $C^3$ is the vector space of ordered triples of complex numbers with the usual addition and scalar multiplication.

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 $\alpha= i$.
I assume that $\alpha=i$.
$\alpha(\xi_1,\xi_2,\xi_3)=(\alpha\xi_1,\alpha\xi_2 ,\alpha\xi_3),\;\xi_1\in\mathbb{R}$.
The set $\mathbb{V}$ is not closed under scalar multiplication and is therfore not a vector space.
Is this better?

Originally Posted by HallsofIvy
(b)
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?
If $\xi_1=0$, the set consisting of vectors of the form $(0,\xi_2,\xi_3)$ is closed under both scalar multiplication and addition. It is a vector space.

Originally Posted by HallsofIvy
(c)
Both (0, 1, 1) and (1, 0, 1) are in this subset. Is their sum?
No it's not. The book lists a couple of axioms valid for a vector space. When I first learned about vector spaces I was told that they are closed under scalar multiplication and vector addition.
Could you tell me which axiom fails here?

Originally Posted by HallsofIvy
(d)
Yes, that is correct. I assume you have shown that.
Now that you have shown me (e), I should be able to show this in a nices way.
Both $(1,-1,1)\;and\;(-1,1,1)$ are in the subset. Their sum is also in the subset.

Originally Posted by HallsofIvy
(e)
Again, both (1, 0, 1) and (0, 1, 0) are in this subset. Is their sum?
Their sum is not in the subset.

Thanks mate!