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In Hoffman's book, a subspace of a vector space is defined as follows:

Let $V$ be a vector space over the field $F$. A subspace of $V$ is a subset $W$ of $V$ which is itself a vector space over $F$ with the operation of vector addition and scalar multiplication on $V$.

I want to understand this definition more precisely, especially, trying to figure out whether the addition and scalar multiplication is on $V$ or $W$ when defining a subspace $W$ from an (existing) vector space $V$. With the notation of the addition $+_V: V \times V \rightarrow V$ and scalar multiplication $\cdot_{F, V}: F \times V \rightarrow V$, the definition is paraphrased as follows:

Given a vector space $(V, +_V, \cdot_{F, V})$ where $+_V$ and $\cdot_{F, V}$ are the addition and scalar multiplication on $V$, if a subset $W$ of $V$ is a vector space with $+_V, \cdot_{F, V}$, then we say $W$ is a subspace of $V$.

Question: Does the paraphrase precisely convey the idea of the original definition?

Some thoughts:

  1. If the paraphrase is correct, can we also say $R^n$ with operations $+_{C^n}$ and $\cdot_{R, C^n} $ is a vector space? This looks a bit odd to me since we usually consider $+_{R^n}$ and $\cdot_{R, R^n}$ when defining a $R^n$ vector space. But looking up various resources regarding the operations within the definition of vector space, it seems there is no restriction for using an operation defined for a larger set (e.g., $C^n$) on a smaller (target) set (e.g.,$R^n$)

  2. If the paraphrase is wrong, should we simply correct it as follows?

    Given a vector space $(V, +_V, \cdot_{F, V})$ where $+_V$ and $\cdot_{F, V}$ are the addition and scalar multiplication on $V$, if a subset $W$ of $V$ is a vector space with $+_W, \cdot_{F, W}$, then we say $W$ is a subspace of $V$. subspace of $V$."

    The only change from the previous paraphrase is "...vector space with $+_W, \cdot_{F, W}$...". But I don't think this is correct because $+_W, \cdot_{F, W}$ are not defined in the Hoffman's definition, and they are different from $+_V, \cdot_{F, V}$. Also, even they are defined, the paraphrase is not aligned with Hoffman's definition, in which the operations are said to be defined on $V$, i.e.,

    "...with the operation of vector addition and scalar multiplication on $\textbf{$V$}$."

Ryan
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  • $\mathbb{R}^n$ is not a vector space with the restrictions of the operations of $\mathbb{C}^n$ because it is not closed under scalar multiplication: the image of $\cdot_{\mathbb{C}^n\mathbb{R}}$ (using your notation) is not contained in $\mathbb{R}^n$. – Arturo Magidin Aug 21 '23 at 05:28
  • @ArturoMagidin Thanks for pointing out. I have corrected the typo – Ryan Aug 21 '23 at 05:33
  • @ArturoMagidin It will be really appreciated if you could share any thoughts on the paraphrase or the ambiguity of (the domian/codomain of) the operations when defining a subspace. Thank you – Ryan Aug 21 '23 at 05:47
  • Your paraphrase can't be right. A vector space $X$ has an addition map $+_X:X\times X\to X$. If you recall the technical fact that this means that $+_X$ is a certain sort of subset of $(X\times X)\times X$, what your book is saying in an informal (but clear way) is that $W$ is a vector space with the addition $+_W:=+_V\cap (W\times W)\times W$. Ditto for the scalar multn. And your paraphrase has failed to capture the fact implicit in the original that $0_V\in W$. – ancient mathematician Aug 21 '23 at 06:42
  • @ancientmathematician Thanks for the comment! I think you are referring a theorem that we can use to determine if a subset $W$ of a vector space $V$ is a subspace of $V$. But the question is more towards the fineness of the definition of subspace itself. In the definition of subspace, it's more like if some properties hold for $W$, then we give it a name - a subspace of $V$. More importantly, one of the property is that $W$ is a vector space. In other words, we do not determine if $W$ is a vector space for the definition of subspace; it's one of the condition/assumption. – Ryan Aug 21 '23 at 07:28
  • @ancientmathematician please refer to this link: https://www.cfm.brown.edu/people/dobrush/cs52/Mathematica/Part3/subspace.html. It gives the definition of subspace (similar to the one in Hoffman's book) and a theorem to determine if a subset of a vector space is a subspace. My question in the post is for the definition itself. Thanks – Ryan Aug 21 '23 at 07:34
  • No, I am not referring to the Subspace Test theorem. I am giving you an exact mathematical formulation of what your book's informal statement means. – ancient mathematician Aug 21 '23 at 07:38
  • @ancientmathematician thanks for the reply! "A vector space $X$ has an addition map $+_X: X \times X \rightarrow X$" is indeed the key to the question. And I find it hard to verify this exactly in various resources, in which the operations are not defined explicitly in mathematical formulation. If we construct a new addition operation $+_Y$ such that $+_X$ is its subset, can we still claim the set $X$ (with $+_Y$ as its addition map and the unchanged scalar multiplication) a vector space? – Ryan Aug 21 '23 at 08:28
  • I'm sorry, I don't understand your question - what's $Y$. The crucial thing (I think) is this. For every subset $X$ of a vector space $V$ one can define the set $+_X$ as I did in my first comment. But in many cases $+_X$ fails to be a function defined on all of $X\times X$ and so $X$ fails to be a subspace. – ancient mathematician Aug 21 '23 at 08:36
  • @ancientmathematician Thanks! I think the main confusion of the definition of subspace is not from the concepts related to subspace, but the definition of the operations for vector space. Say if we have two vector spaces $X$ and $Y$, with the relationship that $X \subset Y$ and $+_X \subset +_Y$. Is the set $X$ (with $+_Y$ as its new addition operation and $\cdot_X$) a vector space? ($\cdot_X$ is $X$'s original scalar multin) – Ryan Aug 21 '23 at 08:55
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    $+_Y$ can't be the addition on $X$ because it's not defined on $X$ but on a superset of $X$. If we're going to be careful about our definitions then we've got to be careful! – ancient mathematician Aug 21 '23 at 11:12
  • @ancientmathematician Thanks for the clarification. Really appreciate your help! – Ryan Aug 22 '23 at 07:02

2 Answers2

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The paraphrase is attempting to be more rigorous than the given definition, but it is not rigorous enough, so you are not really resolving your confusion.

Let's start with something a bit simpler: suppose you have sets $A$ and $B$, and a function $f\colon A\to B$. Given any subset $A'\subseteq A$, you can define the restriction of $f$ to $A'$, often denoted $f|_{A'}$. This is a function $f|_{A'}\colon A'\to B$ whose definition if $f|_{A'}(a') = f(a')$ for every $a'\in A'$. That is, the restriction of $f$ differs from $f$ by virtue of having a different domain, but othwerwise takes the same values.

If $g\colon A\to B$ is a function, $B'$ is a subset of $B$, and $g(A)\subseteq B'$, we can also define the co-restriction of $g$ to $B'$ (I am not aware of any standard or common notation for this, however), as $g'\colon A\to B'$, where $g'(a)=g(a)$ for every $a\in A$. When you care about both domain and codomain of a function, this is a function that is different from $g$ because it has different codomain.

If $\mathbf{V}$ is a vector space over $F$, then vector addition is a binary operation on $\mathbf{V}$, and hence a function $+_{\mathbf{V}}\colon\mathbf{V}\times\mathbf{V}\to\mathbf{V}$. We usually use infix notation for operations, rather than the traditional (in the United States, at any rate) prefix notation for functions. So we write "$\mathbf{x}+_{\mathbf{V}}\mathbf{y}$" to mean the result of applying the function $+_{\mathbf{V}}$ to $(\mathbf{x},\mathbf{y})$. That is, instead of writing the result of using the function as $+_{\mathbf{V}}(\mathbf{x},\mathbf{y})$, we write $\mathbf{x}+_{\mathbf{V}}\mathbf{y}$. Similarly, scalar multiplication is a function $\cdot_{F,\mathbf{V}}\colon F\times\mathbf{V}\to\mathbf{V}$, and we usually denote $\cdot_{F,\mathbf{V}}(\alpha,\mathbf{x})$ by $\alpha\cdot_{F,\mathbf{V}}\mathbf{x}$, or just $\alpha\mathbf{x}$.

If $\mathbf{W}$ is a subset of $\mathbf{V}$, then $\mathbf{W}\times\mathbf{W}$ is a subset of $\mathbf{V}\times\mathbf{V}$, so we can consider the restriction of $+_{\mathbf{V}}$ to $\mathbf{W}\times\mathbf{W}$. And since $F\times\mathbf{W}$ is a subset of $F\times\mathbf{V}$, we can consider the restriction of $\cdot_{F,\mathbf{V}}$ to $F\times\mathbf{W}$.

Suppose that we do that, and we have that whenever $\alpha\in F$, $\mathbf{x},\mathbf{y}\in \mathbf{W}$, we also have $\alpha\cdot_{F,\mathbf{V}}\mathbf{x}\in\mathbf{W}$ and $\mathbf{x}+_{\mathbf{V}}\mathbf{y}\in\mathbf{W}$. Then we can also consider the co-restrictions of $+_{\mathbf{V}}$ and $\cdot_{F,\mathbf{V}}$ to $\mathbf{W}$. Let us write this as $\overline{+_{\mathbf{V}}|_{\mathbf{W}\times\mathbf{W}}}$ (the co-restriction of the restriction to $\mathbf{W}\times\mathbf{W}$, and $\overline{\cdot_{F,\mathbf{V}}|_{F\times\mathbf{W}}}$.

With me so far? Okay, then the definition of "subspace" says:

If $(\mathbf{V}. +_{\mathbf{V}}, \cdot_{F\mathbf{V}})$ is a vector space over $F$, and $\mathbf{W}\subseteq \mathbf{V}$ is a subset of $\mathbf{V}$, then we say that $\mathbf{W}$ is a subspace of $\mathbf{V}$ if and only if $(\mathbf{W},\overline{+_{\mathbf{V}}|_{\mathbf{W}\times\mathbf{W}}}, \overline{\cdot_{F,\mathbf{V}}|_{F\times\mathbf{W}}})$ is a vector space over $F$.

(And the clutter probably lets you see why we usually try to paraphrase this rather than make the notation be very accurate...)

About your thoughts:

  1. No. Note that while the restriction of $+|_{\mathbf{C}^n}$ to $\mathbb{R}^n$ does work (meaning, you can also do the co-restriction), the same is not true of $\cdot|_{\mathbb{C},\mathbb{C}^n}$: if you try to restrict it from $\mathbb{C}\times\mathbb{C}^n$ to $\mathbb{C}\times\mathbb{R}^n$, you can no longer also co-restrict it: for instance, $i\cdot (1,1,\ldots,1)$ does not lie in $\mathbb{R}^n$. So this restriction is not a "scalar multiplication" on $\mathbb{R}^n$. However, if you consider $\mathbb{C}^n$ as a vector space *over $\mathbb{R}$, then you do get that $\mathbb{R}^n$ is a subspace of $\mathbb{C}^n$ over $\mathbb{R}$.

  2. The problem with your proposal in item 2 is that you do not explain what $+|_{\mathbf{W}}$ and $\cdot|_{F,\mathbf{W}}$ mean. You would need to say something like "where by $+|_{\mathbf{W}}$ we mean blah", etc.

The idea of the definition is that $\mathbf{W}$ should be a vector space in its own right, when you add vectors in $\mathbf{W}$ exactly the same way as you add them when you consider them as elements of $\mathbf{V}$. This is analogous to how when we add two integers as if they were fractions we get the same answer as when we add them as integers. We usually think of addition of rationals as extending the addition of integers (we already know how to add integers, we are not using that to explain how to add rationals). But if for some reason we knew how to add rationals but not how to add integers, we could ask whether when you add two fractions-correspond-to-integers we get a fraction-that-corresponds-to-an-integer, and if so restrict the addition-of-rationals to integers to get a way to add integers.

Arturo Magidin
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  • Thanks so much for your insightful comment! This really provides me a better perspective on domain/codomain of a function and the definition of subspace. As it was mentioned that my paraphrase is not rigorous enough but is still correct, this makes me wonder if the following is correct: for any vector space $(W, +W, \cdot{F, W})$ over $F$, if there exists a vector space $(V, +V, \cdot{G, V})$ over $G$ such that $W \subset V$, $+W \subset +_V$ and $\cdot{F, W} \subset \cdot_{G, V}$, then $(W, +V, \cdot{G, V})$ is also a vector space? – Ryan Aug 22 '23 at 04:49
  • @Ryan the way you write it, no. Because if you are going to define a vector space as an ordered triple were the first entry is a set, the second entry is a binary operation on the set in the first entry, the third entry is a blah, such that they satisfy the relevant properties (vector space axioms), then $(W, +V,\cdot{F,W})$ cannot be a vector space because the second entry of the triple is not the correct type of object: it is not a binary operation on $W$, as it is not a function $W\times W\to W$. – Arturo Magidin Aug 22 '23 at 04:51
  • @Ryan Also, the change in fields screws you up. (By the way, we usually use $F$, $K$, $L$ for fields, never $G$). It may be impossible to extend scalar multiplication. E.g., there is no way to make $\mathbb{R}^1$ into a complex vector space if you want the scalar multiplication by reals to be the usual one. – Arturo Magidin Aug 22 '23 at 05:06
  • Thanks! So in this way, I think maybe my paraphrase is incorrect. The reason is for any set $W$ to be a vector space, the addition operation is required to be a binary operation on $W$, i.e, $+W: W \times W \rightarrow W$; but in my paraphrase, it says "...a subset $W$ of $V$ is a vector space with $+_V$ and $+{F, V}$...", which contradicts to the definition of a vector space (for which the addition operation needs to be defined on $W$) even though we have $W \subset V$ and $+_W \subset +_V$. Please correct me if I am wrong. Thanks – Ryan Aug 22 '23 at 05:10
  • @Ryan That sounds accurate. That's why I had to take the co-restriction and the restriction in the definition I give. What one does in practice is take just the restriction $+|_{W\times W}\colon W\times W\to V$, and then check that the image is contained in $W$, and then simply elide that you can then take the co-restriction to formally define the new operation. – Arturo Magidin Aug 22 '23 at 05:19
  • That co-restriction and the restriction really clarify everything. Thanks so much for your help! Really enjoy the discussion :) – Ryan Aug 22 '23 at 05:29
  • @ArturoMagidin There seems to me to be one detail not yet dealt with. I think that as you define "subspace" the empty set would qualify, which it oughtn't. So either we need to insist that $W\not=\emptyset$ or (as I prefer) include the $0$-ary operator which picks out the zero vector to our definition of vector-space. – ancient mathematician Aug 22 '23 at 08:17
  • @ancientmathematician The definition is that $W$, with the restricted operations, be a vector space. Since "is a vector space" requires the existence of a zero vector in the set, you are incorrect. – Arturo Magidin Aug 22 '23 at 13:18
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    @ancientmathematician I think you are confusing "$W$ is a vector space under the restricted operations" (the given definition) with "$W$ is closed under the operations of $V$ (which is not what is stated). – Arturo Magidin Aug 22 '23 at 13:25
  • @ArturoMagidin Thanks. That clarifies it for me. – ancient mathematician Aug 22 '23 at 13:54
  • @ArturoMagidin Can't one get away without using the word "restricted": "W is a vector space under the operations of V", further reducing verbal clutter :) – rych Aug 23 '23 at 04:11
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    @rych which is the phrasing that caused the OPs confusion to begin with... thereby reducing "clutter" at the expense of understanding. Seems like a terrible exchange. – Arturo Magidin Aug 23 '23 at 05:39
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Question: Does the paraphrase precisely convey the idea of the original definition?

Yes.

The addition and scalar multiplication on the elements of the candidate subset are carried out in the original, ambient vector space. If all results fall again into the candidate subset, it is a subspace.

This is correctly mentioned in the Wikipedia as if under the operations of V, W is a vector space. The accepted answer above includes an interesting discussion rephrasing the same idea as domain restriction and co-restriction. But basically, the definition of subspace does not introduce any new operations.

rych
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  • Thanks for your input! I think the essence of the question is more related to how binary operations should be defined in the definition of vector space. E.g., If the following is true, then introducing new addition operation is not necessary for defining a subspace: "for any vector space $(W, +W, \cdot{F, W})$ over $F$, if there exist a set $V$ and an addition operation $+V$ such that $W \subset V$ and $+_W \subset +_V$, then $(W, +_V, \cdot{F, W})$ is a vector space over $F$" – Ryan Aug 22 '23 at 07:37
  • Or simply if the addition operation for a vector space $W$ is required to be defined as $+_W: W \times W \rightarrow W$. If yes, then we have no other way but using the idea of restriction (and co-restriction) for subspace definition. – Ryan Aug 22 '23 at 07:48
  • @Ryan, you could say that restriction takes place here either implicitly or explicitly. When we apply a binary operation over a subset, do we "restrict" it to the subset? When we confirm the result belongs to the same subset do we "restrict" the co-domain? Allow me to stand by my answer, which is also Wikipedia answer, and to the phrase "if under the operations of V, W is a vector space" let me just add "over the same field"! – rych Aug 22 '23 at 10:23
  • Thanks. Yea..as I have mentioned it really depends on what is allowed in defining the operation(s) in the definition of vector space, i.e., for any set $W$ to be a vector space, are we allowed to use an addition operation $+V$ on $W$, where $V$ is the superset of $W$ and $+_V$ (when suitably restricted) is closed under $W$. An _explicitly answer on this will directly address the question/ambiguity (of definition of subspace) in the post. – Ryan Aug 22 '23 at 20:43