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According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace $V$ is defined to be the minimal number of vectors required to generate $V$ (pp. 4).

Then why does $\{\mathbf{0}\}$ have dimension zero instead of one? Shouldn't it be true that only the empty set has dimension zero?

Qingtian
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    The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space. – Michael Joyce Feb 05 '14 at 13:45
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    it is a definition – Semsem Feb 05 '14 at 13:57
  • @MichaelJoyce see this http://math.stackexchange.com/questions/1315457/has-the-point-dimension-zero-or-one – user75086 Jun 07 '15 at 08:55
  • Possibly helpful thought: a countable set has (Hausdorff) dimension zero (as a subset of $\mathbb R^n$). The "volume" spanned by the zero vector is zero. Of course if you take a discrete space as the "parent" then it can be different. – jdods Aug 06 '16 at 13:42

4 Answers4

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A vector by itself doesn't have a dimension. A subspace has a dimension. Why $\{\mathbf{0}\}$ is considered as having dimension $0$? Because of consistency with all other situations. For instance $\mathbb{R}^3$ has dimension $3$ because we can find in it a linearly independent set with three elements, but no larger linearly independent set. This applies to vector spaces having a finite spanning set and so of subspaces thereof.

What's the largest linearly independent set in $\{\mathbf{0}\}$? The only subsets in it are the empty set and the whole set. But any set containing the zero vector is linearly dependent; conversely, the empty set is certainly linearly independent (because you can't find a zero linear combination with non zero coefficients out of its elements). So the only linearly independent set in $\{\mathbf{0}\}$ is the empty set that has zero elements.

egreg
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The sum over the empty set is the additive identity... in this case zero.

JP McCarthy
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If $~0~$ is a part of your basis, and if it is the only vector, there are zero linearly independent vectors in it, since any non-negative multiple of $~0~$ is $~0~$.

Thus, the dimension of the space is $~0~$.

nmasanta
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Laukik Mujumdar
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Short answer: Because its basis is the empty set $\emptyset$.

If $V$ is a set with exactly one element and $F$ is a field, there is exactly one way to define addition and scalar multiplication such that $V$ is a vector space over $F$. In this case, $\emptyset$ is the only linearly independent subset. Since the span of $\emptyset$ - the intersection of all vector subspaces containing $\emptyset$ - is equal to $V$, $\emptyset$ is even a basis - the basis of $V$.

Filippo
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  • Absolutely any single vector space has a basis, there is no such thing as a vector space without a basis....A null vector space has a unique basis, namely the empty set which is nevertheless a basis. – ΑΘΩ Apr 22 '21 at 01:24
  • @ΑΘΩ Here's what I think: A basis of $V$ is a subset of $V$ that is linearly independent and a generating set. The empty set is a linearly independent subset of all vector spaces, but not a generating set and hence not a basis. Do you agree? – Filippo Apr 22 '21 at 07:23
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    I do not agree for I cannot, as the claim that the empty set is not a generating set is wrong. By definition, given a (left) vector space $V$ over a certain field $K$ and an arbitrary subset $X \subseteq V$, the $K$-subspace generated by $X$ is the intersection of all subspaces of $V$ which include $X$. This intersection always comprises at least one factor ($V$ itself) and turns out to be the minimum (with respect to inclusion) subspace of $V$ which includes $X$. Applying this definition to $X=\varnothing$ leads to the immediate conclusion (to be cont.) – ΑΘΩ Apr 22 '21 at 10:02
  • (cont.) that the subspace generated by $\varnothing$ is the null subspace ${0_V}$ (the set of all $K$-subspaces of $V$ which include $\varnothing$ is equal to the set of all $K$-subspaces whatsoever and the null subspace obviously is its minimum). In particular it follows that as long as $V$ itself is null it is generated by the empty set. – ΑΘΩ Apr 22 '21 at 10:05
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    Upon more careful reading of your answer another thing emerged that I would like to point out: the final clause is not very accurate, since whatever proofs we have available to show that vector spaces with at least $2$ elements have a basis also naturally apply to the other kind of vector spaces, namely the null ones. Making a distinction by this criterion and giving separate treatment for this issue (existence of bases) is not at all needed and extremely artificial... – ΑΘΩ Apr 22 '21 at 10:15
  • @ΑΘΩ Thank you very much for your explanation! What I wrote in my answer is what I found in the lecture notes of my professor - but I understand that you are right! I think the problem is that my professor defined the span of $X$ as the set of all linear combinations, which is equivalent to your definition as long as $X$ is nonempty. But in the case of the empty set, your definition is the correct one! Thus, the null subspace has exactly one basis - the empty set. I have edited my answer- thank you for correcting my mistake! – Filippo Apr 22 '21 at 13:10
  • @Filippo : the empty sum is well defined it is $0$. You need this fact to root the recursions needed to give a careful definition of what $\sum_{i=1}^{n}\lambda_i v_i$ means, to give meaning to "linear combinations". – ancient mathematician Apr 22 '21 at 13:58
  • @ancientmathematician Interesting! I guess we could consider the map $S\colon{A\subset V:A\text{ is finite}}\to V$ assigning to each finite subset $A$ of $V$ its sum $S(A)$ (with $S(\emptyset):=0$) to make a precise definition... – Filippo Apr 22 '21 at 14:43
  • @Filippo I am glad that you found my (rather long) comments instructive. I was expecting somehing of the kind to be the background issue and may I make a further remark by saying that what your professor apparently introduced as the definition for subspaces generated by subsets is not actually the appropriate definition but merely an equivalent characterisation. The only natural definition is the one I mentioned in my previous batch of comments. Nevertheless, as ancientmathematician pointed out, (to be cont.) – ΑΘΩ Apr 23 '21 at 11:29
  • @Filippo (cont.) the characterisation of $\langle X \rangle_K$ as being the subset of all linear combinations of the form $\displaystyle \sum_{t \in X}\lambda_t t$ - where $\lambda \in K^{(X)}$ is a family of finite support - is equally valid in the case $X=\varnothing$: in this case $K^{(\varnothing)}=K^{\varnothing}={\varnothing}$, in other words there is a unique family of coefficients of finite support and the linear combination it yields is the empty sum in $V$, in other words none other than $0_V$, by definition of finite sums over (commutative) monoids. – ΑΘΩ Apr 23 '21 at 11:33
  • @ΑΘΩ $K^{(X)}$ is the set of all maps from $X$ to $K$ that assign $0$ to all but a finite amount of elements of $X$? – Filippo Apr 23 '21 at 13:02
  • And $\langle X\rangle={\sum_{x\in X}\lambda(x)\cdot x:\lambda\in K^{(X)}}$? – Filippo Apr 23 '21 at 13:07
  • If that's correct, I really like this definition, because there's no need to talk about "empty sums" - since the map assigning $0\in K$ to each element of $X$ is obviously an element of $K^{(X)}$ and $\sum_{x\in X}\lambda(x)\cdot x=0\in\langle X\rangle$. – Filippo Apr 23 '21 at 13:12
  • @Filippo Not of maps but of families, distinction which is very important as maps will be distinct as long as they have distinct codomains even though they might have the same underlying functional graphic. Also important to note that even in this formulation - I should actually say all the more so in this formulation - empty sums very much do come into play, since any linear combination over an empty $X$ leads to the sum of an empty family, i.e. a family indexed by $\varnothing$. – ΑΘΩ Apr 23 '21 at 15:03
  • @ΑΘΩ You are right, I got confused. We still have to define the expression $\displaystyle \sum_{t \in X}\lambda_t t$, where $\lambda\colon\emptyset\to K$, separately. That's why I like your other definition of the span better. – Filippo Apr 23 '21 at 17:18
  • Let us continue this discussion in chat. – ΑΘΩ Apr 24 '21 at 04:33