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If we have a Vector Space such as $\Bbb R$, we can make Vector Spaces out of it. For example, let $v_1,v_2,v_3\in \Bbb R$. We know that $(v_1,v_2,v_3)$ is a Vector Space - it is $\Bbb R_3$ - and its dimension is $3\times \dim {(\Bbb R)} = 3$. Thus we can see that $\Bbb R_3$ is composed of three elements of $\Bbb R$.

Any "Container Space" ($V_1$) is made of a Vector Space of elements whose elements are vectors in another Vector Space ($V_2$). Also, $(v_1,v_2,...,v_n) \oplus_1 (v'_1,v'_2,...,v'_n) = (v_1\oplus_2v'_1,v_2\oplus_2v'_2,...,v_n\oplus_2v'_n)$ and $c\odot_1(v_1,v_2,...) = (c\odot_2 v_1, c\odot_2 v_2...)$. In other words, Operators are passed to the component vectors.

In general, if we have a Vector Space $V$ and a Set $C$ made from vectors in $V$, is $C$ always a Vector Space? If so, will the dimension of $C$ always be $\dim{(P)}\times\dim{(V)}$ where $P$ is the space that forms elements in the way that the Set $C$ does?


I have found a few more Vector Spaces of the same type; if you want more examples, I can post them.

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  • $\begingroup$ The second paragraph is confusing... I don't see what you mean by "parent space". What you literally written is something like "The parent space is an element of $R_n$ and matrices with entries from $R$." I have no idea what you intend by this. Can you reexplain? $\endgroup$ – rschwieb May 16 '13 at 21:02
  • $\begingroup$ Look at what you've written: "If we have a vector space $V$ and a vector space $C$... is $C$ a vector space?" This does not make sense: you are assuming things and then asking questions about the assumptions. $\endgroup$ – rschwieb May 16 '13 at 21:09
  • $\begingroup$ @rschwieb What am I assuming? Where are the assumptions? $\endgroup$ – Justin May 16 '13 at 21:17
  • $\begingroup$ You say "If we have... a vector space $C$... is $C$ a vector space?" You assume $C$ is a vector space and then ask if it is a vector space. $\endgroup$ – rschwieb May 16 '13 at 21:20
  • $\begingroup$ What you describe in your second paragraph is exactly the direct product (or direct sum) of $n$ vector spaces. $\endgroup$ – rschwieb May 16 '13 at 21:24
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Wait a moment. I think there are some things confused here. Given $(v_1,v_2,v_3)$ for each $v_i \in \mathbb{R}$ this tuple isn't a vector space, rather it's the element of another vector space, namely $\mathbb{R}^3$. So, how do we construct new vector spaces from old ones? Well, there are two methods that I think are very interesting, and that probably will interest you.

The first method which is the simpler (and which is what you are looking for, looking at your question) is the so called Direct Sum. Given a family of vector spaces $\{V_i : i \in I\}$ we define the direct sum as the result of taking the cartesian product of the $V_i$ and adding the operations componentwise as defined in each $V_i$. In other words, elements of the direct sum are of the form $(v_1, \dots, v_n)$ with each $v_i \in V_i$ and we define the operations componentwise in the sense that we have:

$$(v_1,\dots, v_n)+(w_1,\dots,w_n)=(v_1+w_1,\dots,v_n+w_n)$$

$$a(v_1,\dots, v_n)=(av_1,\dots,av_n)$$

Note that the operation in the $i$-th component is the operation defined in the space $V_i$. This vector space is denoted by the symbol $V_1 \oplus \cdots \oplus V_n$ or simpler with the notation:

$$\bigoplus_{i\in I}V_i$$

It's fairly easy to prove that $\dim(\oplus_{i\in I}V_i) = \sum_{i \in I}\dim V_i$ (and I invite you to do that). This is the way $\mathbb{R}^3$ is construct in truth. You can see that with this definition we have:

$$\mathbb{R}^3 = \bigoplus_{i=1}^3\mathbb{R}$$

The other method is called the tensor product. It's construction is kind of sofisticated (and in truth it's motivation is only clear when we come to study multilinear constructions), but in the end it's simple when you understand it: it's meant to mimic usual products. This space is denoted by $V_1 \otimes \cdots \otimes V_n$ or simpler:

$$\bigotimes_{i \in I}V_i$$

And it satisfies $\dim(\otimes_{i\in I}V_i) = \prod_{i\in I}\dim V_i$ so in some intuitive sense the tensor product is bigger than the direct sum. To read more about the construction of the tensor product consult Kostrikin's Linear Algebra and Geometry chapter 4 about Multilinear Algebra.

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  • $\begingroup$ There are certainly a lot of things confused. I can't tell if by $R$ the poster meant $\Bbb R$ or not. $\endgroup$ – rschwieb May 16 '13 at 21:12
  • $\begingroup$ Well, now reading the question again I realize I didn't understand what the OP wanted. I think that he want to consider $\mathcal{A}$ a family of vector spaces and try to make this thing a vector space. I'm not sure. $\endgroup$ – user1620696 May 16 '13 at 21:16
  • $\begingroup$ I know what you mean: the user is not conveying what he means very well at this point. $\endgroup$ – rschwieb May 16 '13 at 21:24
  • $\begingroup$ @rschwieb I'm sorry about that, I attempt to make my intentions clear, but sometimes I fail. $\endgroup$ – Justin May 16 '13 at 21:26
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"Piano tiger", all of these things have normal names, which would be much more preferable to use than these new ones you've invented. (Actually, I'm not even sure two of those names are necessary to invent, unless your intention is to boggle the reader.)

Given a vector space $V$, you can always form the product $V\times V\times V$ (which you could choose to denote as $V^3$). This is the product space of three copies of the space $V$.

The dimension of the product is always the sum of the dimensions of the pieces of the product... so if $\dim(v)=n$, then $\dim(V^3)=3n$.

In general, $\dim(V^k)=k\cdot \dim(V)$.


Rather than saying "$C$ is a container space of $S$", it would be more conventional to say that $S$ is a subspace of $V$.


Your latest revision of your third paragraph can now be translated as: "If $P$ is a subspace of $V$, and $C$ is a set made from vectors of $V$, is $C$ a vector space?"

If $C$ is just some random subset of $V$, then no, it is unlikely that it is a vector space. For one thing, it might not have the zero vector.


Latest addition for latest edition.

The set you described forming from elements of $P$ is just another product space $P^n$. That $P$ is a subspace of $V$ does not change anything about forming the product... it certainly does not need a newly invented name. A $n\times n$ matrix with entries from $V$ is just a fancy way of writing a vector of length $n^2$. It is the same as $V^{n^2}$ except for the way you write it.

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  • $\begingroup$ This? $\endgroup$ – Justin May 16 '13 at 21:00
  • $\begingroup$ No, I didn't say inner product, I meant this $\endgroup$ – rschwieb May 16 '13 at 21:01
  • $\begingroup$ Wikipedia can be hard to read if you don't know what symbols mean. Sadly, I don't know what too many symbols mean. I see that the Direct Product is composed of elements of another vector space, but how does the $+$ and the $\bullet$ work? $\endgroup$ – Justin May 16 '13 at 21:08
  • $\begingroup$ @gangqinlaohu Well it turns out if you are only using finitely many copies of $V$, then $V\times V$ and $V\oplus V$ mean the same thing. They are just vectors of vectors, as you described. $\endgroup$ – rschwieb May 16 '13 at 21:10
  • $\begingroup$ vectors of vectors... don't encourage him. $\endgroup$ – Narut Sereewattanawoot May 16 '13 at 21:25

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