I have a field $F$ and $n \in \mathbb{N}$.

For any $a \in F$, $V_a$ is a defined as:
$V_a := \{(v_1,v_2,…,v_n) \in F^n | v_1+v_2+…+v_n = a\}$

The question is for which $a \in K$ is $V_a$ a vector subspace of $F^n$ ?
I should then calculate the basis and dimension of these $V_a$.

I obviously don't want you to simply give me the solution to this problem, but instead tell me if I did it right and maybe give me tips on what's wrong / what I can improve on. So here's what I did:

$V_0$ is a vector subspace for every $n$:

$\forall v \in V_0: v_1 + … + v_n = 0$
$α(v_1,…,v_n) = (αv_1,…,αv_n) \in V_0$, since
$αv_1 + … + αv_2 = α(v_1 + … + v_n) = α*0 = 0$

$\forall v_1,v_2 \in V_0: (v_{1,1},…,v_{1,n}) + (v_{2,1},…,v_{2,n}) = (v_{1,1} + … + v_{1,n}) + (v_{2,1} + … + v_{2,n}) = 0 + 0 = 0$

However, if $a≠0$, the addition of two vectors from $V_a$ does not have to be in $V_a$ again, so it's not a vector space on its own and can therefore not be a vector subspace of $F^n$.

Is this correct?

For $V_2$ the result would be a line like $y=-x$, for $V_3$ a plane like $x+y+z=0$, etc.
So I thought that $dim(V_n) = n-1$. However, I have no idea if that's correct and how to prove that, and also what $dim(V_1)$ would be.

Also, how do I calculate a corresponding basis for these $V_n$ ?

Last but not least, what would a base of the quotient space $F^n/V_0$ look like?

  • $\begingroup$ For $ V_a $ to be a subspace of $ F^n $ it must contain the zero vector of $F^n $. This fact alone implies that $ V_0 $ is your only candidate. $\endgroup$ – DKal Dec 9 '13 at 14:43

I think between your question and your last paragraph you confuse what the index $a$ of $V_a$ stands for. The examples you show (i.e. the plane $x + y + z = 0$) are all $V_0$, in the corresponding $\Bbb R^n$.

Because of scalar multiplication, you need $ax = a$ for all $x \in F$ if $V_a$ is to be a vector space, and that is only statisfied for $a = 0$. You have shown that $V_0$ is a vector space, so you have classified all vector spaces of the form $V_a$.

For the dimension, note that to construct a point in $V_a$ (it still has a dimension, even though it's not necessarily a vector subspace, you'll get to that in algebraic geometry), you can freely choose the first $n-1$ coordinates. The last coordinate will then necessarily be fixed. This is not necessarily a rigorous proof that the dimension is $n-1$ (to do that you first need to take a good look at your definition of dimension), but it should be enough to convince you that it is no coincidence that it works for $F = \Bbb R$ and $n = 1, 2$ and $3$.


Hint: You know that vectors of the form $(0,\ldots, 0,1,0,\ldots,0,-1,0,\ldots,0)$ are in your vector subspace.


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