0
$\begingroup$

I have questions about checking whether or not a space is a vector subspace of $E$, a $n$-dimensional real vector space.

I understand the rules for defining a vector space (commutativity, distributivity, stability under linear combination, presence of $0$ and so on), but I can't really figure out how to apply these when presented such case:

if $E = R^n$ , check that $F = \{ x = ( x_1 ,\dots, x_n ) : \sum_{i=1}^n x_i = 0 \}$ is a vector subspace of $E$.

Also check case of $G = \{ x = ( x_1 ,\dots, x_n ) : \sum_{i=1}^n x_i = 1 \}$

My problem is that $x_n$ are vectors in the $n$-th dimension, so how can they sum up to a scalar. Also, how can I test a singleton (like $\{0\}$) presence's for such a sum?

For what i understand of this, in any case, any $x_n$ could be a $\{0\}$ in its own dimension and not change anything to the sum.

Also, I know actually the answers ($F$ is a subspace and $G$ not, I can relate it to the presence of $0$ and the fact that for a vector space, you should have cancelled each vector by an opposite, but yet, I can't understand the notation of a scalar to something that is obviously not one as it is a sum of $n$-dimensional vectors (actually, unidimensional vectors in multidimensional space, so that the $k$-th dimension of vector $(u_1, \dots, u_n)$ is equal to a scalar and the others to $0$.

Would someone enlighten me, please?

$\endgroup$
  • $\begingroup$ Just to clarify, you aren't checking for a singleton $\lbrace 0 \rbrace$, but rather an element $\mathbf{0} \in F$ such that for any $\mathbf{f} \in F$, $\mathbf{0} + \mathbf{f} = \mathbf{f} + \mathbf{0} = \mathbf{f}$. $\endgroup$ – ÍgjøgnumMeg Jun 21 '17 at 14:12
  • 2
    $\begingroup$ The $x_i$ are the scalar components of $x$, not vectors. You need to check that the set contains the zero vector and is closed under linear combinations. Your first set obviously contains the zero vector, because the sum of its components is zero; your second set doesn't contain the zero vector, because the sum of its components isn't 1. So you can conclude the second set is not a subspace. To finish, you need to show the first set is closed under linear combinations. Well, if you add two vectors whose components sum to zero, it's easy to see the components of the result will also sum to zero. $\endgroup$ – symplectomorphic Jun 21 '17 at 14:15
  • 1
    $\begingroup$ The notation $x=(x_1,\ldots,x_n)$ means that $x$ is a vector $\in\Bbb R^n$ and the individual $x_i$ are its components, hence elements of $\Bbb R$. For example, if $n=3$, you might have $x=(3,-4,2)$ and then $x_1=3$, $x_2=-4$, $x_3=2$. As $x_1+x_2+x_3=3-4+2=1$, we see that $x\in G$. $\endgroup$ – Hagen von Eitzen Jun 21 '17 at 14:16
  • 1
    $\begingroup$ And finally, when you scalar multiply a vector whose components sum to zero, you still get components that sum to zero. (Why?) $\endgroup$ – symplectomorphic Jun 21 '17 at 14:16
  • $\begingroup$ From what I get from your answers, $x_i$ are scalars, ok, but then they are subvectors of the vector they compose (since they are scalar that myltiply base vectors from each dimension of R^n ). And actually, I don't see how different dimensions can cancel the other out. Plus, should I understand that F and G only contain one vector? As for the latter question, multiplying by zero gives 0, indeed, but it is so obvious it can't be used as a proof, no operation is more constant than that. $\endgroup$ – Ando Jurai Jun 21 '17 at 14:27
2
$\begingroup$

You're overthinking it. When we say $\mathbb{R}^n$ is the set of all $n$-dimensional vectors $(x_1,\dots, x_n)$, then $x_1,\dots,x_n$ are just scalars. They're not one-dimensional vectors or subvectors or anything like that. As such, you can combine them just like any elements of a field.

For instance, if $n=3$, then $F$ contains the vector $(1,0,-1)$ and $(2,-1,-1)$. This is because $1+0+(-1) = 0$ and $2+(-1)+(-1) = 0$. If you add these two together, you get $(3,-1,-2)$. Notice that this vector is still in $F$. That is some evidence that $F$ might be a subspace.

On the other hand, $G$ contains $(1,1,-1)$ and $(3,-1,-1)$. If you add these together, you get $(4,0,-2)$, which is not in $G$. This means that $G$ can't be a subspace.

In general, if $(x_1,\dots,x_n) \in F$, you know that $x_1 + \dots + x_n = 0$. Is this relation preserved by scalar multiplication? Is the zero vector $(0,\dots,0) \in E$ also in $F$? These are the things you need to show.

$\endgroup$
  • $\begingroup$ Thanks for the explanation. I indeed can accept this idea, but actually we are always shown cartesian plane and sum of geometric vectors in my course to illustrate the statements about x+lambda*y=z and so on. So that's why I don't get the "informational vector" thing. Scalar multiplication of a vector whose component sum up to zero will equal to 0 indeed (that is mere algebra and I indeed understood it, what I have difficulties to cope with is the logical premisses). Also, then we are taught about bound spaces and so on, and you need to sum by individual components of the v_s $\endgroup$ – Ando Jurai Jun 21 '17 at 15:33
  • $\begingroup$ @AndoJurai: When you say you don't get the "informational vector" thing, do you mean you'd prefer to think of a vector as an arrow in space rather than a list of numbers? Fair enough, but there are many useful examples of vector spaces in algebra. I don't know what bound spaces are, so I can't help there. $\endgroup$ – Matthew Leingang Jun 21 '17 at 15:44
  • $\begingroup$ sorry, bound families of vectors vs free families of vectors; Like in "some don't sum up to zero while some do". $\endgroup$ – Ando Jurai Jun 21 '17 at 16:04
  • $\begingroup$ It's actually not about what I prefer, I thought about what you said and actually, while I get "vector as list", that's the fact they are represented in planes that makes me wonder. But I guess that my course confuses easily the vectors (whose respective {i_th} dimension can be represented on some axis of in the field) and their dimensions. Somehow, I begin to see clearer. Thanks to all. $\endgroup$ – Ando Jurai Jun 22 '17 at 10:41
0
$\begingroup$

$\forall v \in V: v=\begin{pmatrix} v_1 \\ \vdots \\ v_n\end{pmatrix}$

So, as $v_i$ is a number $\forall i$, you can sum up those. In fact $v_i$ is just a component of the vector.

$\endgroup$
  • $\begingroup$ Since v is a vector, its components are in different dimensions, right? So how can they add? It's what I don't understand. If they are not in different dimensions, then v is not a vector, it's just a list of numbers to add between themselves. But we are talking about vector spaces, so it doesn't make any sense to me there. $\endgroup$ – Ando Jurai Jun 21 '17 at 15:11
  • $\begingroup$ Coordinates are in the field you are considering. Imagine this: $v=a_1v_1+\dots a_nv_n$ and $a_i$, which represents coordinates, is in $\mathbb{K}$. $\endgroup$ – Alberto Andrenucci Jun 21 '17 at 15:26

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.