# When is the additive identity not the zero vector?

My teacher cryptically mentioned today that the zero vector is not always the additive identity. When asked for clarification I was told "we'll get there".

He did confirm it is always 0 in matrices filled with real numbers, but I can't think of or find any matrix, whether complex or variable or whatever where anything else would work, or where the zero vector wouldn't work.

It might be half a joke to keep me interested, but I'll be a minkeys uncle if it didn't work!

I don't know, any ideas?

• As @Batman explains, the zero vector of a vector space always is the additive identity of this vector space. I suspect your professor could have meant that the zero vector of a vector space could be distinct from the additive identity of some other addition on the same set, or that the addition of the vector space is an "addition" in name only (@mrf gives such an example). Sep 23, 2015 at 20:53
• @darijgrinberg, there was some weird question yesterday about en.wikipedia.org/wiki/Dyadics which give some unusual outcomes. Maybe this is more of the same, as usual I can't read minds. Sep 23, 2015 at 21:43
• thanks guys I can't wait to see what my Professor comes up with, I feel like I should pick an "acepted answer" now , but can't figure out how (if I could I'd pick darij grinberg's comment but will probably go with batmans) Sep 23, 2015 at 22:41
• On any uncountable set $X$ and for any element $x \in X$, one can define a vector space structure on $X$ such that $x$ is the additive identity (the 'zero vector'). This is achieved by a transport of structure which can also be used to construct the example in mrf's answer (use the map $\ln : \mathbb{R}_+ \to \mathbb{R}$). I talk about this concept in this answer and this one. Sep 24, 2015 at 16:34

That depends on what you mean by the zero vector.

If you want, you can consider $\mathbb{R}_+$ (the set of positive real numbers) as a vector space, where you define $x\oplus y = xy$ and $\lambda x = x^\lambda$. Then the "additive identity" is actually $1$ (but should probably be called the zero vector in this strange context).

• Sorry if this is a stupid question but are you saying were in R1, and if so wouldn't landau=x. Ill have to hit this with fresh eyes in the morning, I must be missing something (possibly a lot) Sep 23, 2015 at 21:34
• No, the underlying set is not $\mathbb{R^1}$, just the positive real numbers, and the above formulas define vector addition and scalar multiplication on this set.
– mrf
Sep 23, 2015 at 21:39
• isnt the positive numbers by definition not a vector space because scalar multiplication of negative one makes addition not hold? Can we really just redefine addition and multiplication this way? thank you for your quick reponse, I feel I should take some time to grasp what your saying before responding further. Sep 23, 2015 at 21:42
• The positive reals are not a vector space under the usual operations, but you can check that with the above definitions, it is.
– mrf
Sep 23, 2015 at 21:44
• yeah it's an easy check but it seems like changing the definition of addition changes the definition of a vector space, idk I think I get it but forgive me if it seems a little trivial. My mind went racing with this question and I was kind of hoping there was a different kind of answer. Does redefining the most basic of operations ever yield a useful result in your experience? btw I love the creative answer, your right you can do anything you want but doesn't it compromise the structure needed for useful mathematics Sep 23, 2015 at 22:57

A vector space forms a group under addition:

1) If $a,b$ are vectors, $a+b$ is a vector.

2) $(a+b)+c = a+(b+c)$ for vectors $a,b,c$.

3) $a+0 = 0+a = a$ by definition of the zero vector $0$ (i.e. the zero vector is defined to be an additive identity as in the vector space axioms).

4) $a+(-a) = (-a) + a = 0$ , i.e. additive inverses exist.

In a group, one can easily show an additive identity is unique (hence why we say "the" additive identity).

Thus, the zero vector is the only additive identity.

• Thats what I figured but the other answer makes me and my teacher are hurting my brain tonight! But your answer makes sense, its defined that way for a reason. much more pondering is called for. Also sorry for half stealing your name, if I wouldn't've picked it (but I would still be the real Batman) if Id've known Sep 23, 2015 at 21:29

The difficulty is not to confuse the abstract notion of a vector space with concrete vector spaces. What do I mean by this?

### The notion of an abstract vector space

Well, if $0_\mathbb{V} \in \mathbb{V}$ is a distinguished element of a non-empty set $\mathbb{V}$, $\mathbb{K}$ a field, and $\oplus : \mathbb{V} \times \mathbb{V} \to \mathbb{V}$ and $\otimes : \mathbb{K} \times \mathbb{V} \to \mathbb{V}$ two operation such that $0_\mathbb{V}$, $\oplus$, and $\otimes$ satify the vector space axioms; then we can call the algebraic structure $$(\mathbb{V}, \oplus, \odot, 0_\mathbb{V})$$ an abstract $\mathbb{K}$-vector space. It is abstract in the sense that $\mathbb{V}$, $\mathbb{K}$, $\oplus$, $\odot$, and $0_\mathbb{V}$ are placeholders for concrete mathematical objects. Nonetheless, we want to be able to talk to others about the different placeholders, which is why we agree upon calling the symbol $\oplus$ vector addition, the symbol $\odot$ scalar multiplication, and the symbol $0_\mathbb{V}$ additive identity.

Supplement: To show how important it is to keep the different notations apart, consider the linear combination $$v = k_1 \odot v_1 \oplus k_2 \odot v_2 \oplus \ldots \oplus k_n \odot v_n = \bigoplus\limits_{i=1}^{n} k_i \odot v_i.$$ Then we call the symbol $\bigoplus$ vector summation.

### A concrete vector space 1

A standard example of a concrete $\mathbb{R}$-vector space is $(\mathbb{R}, +, \cdot, 0)$. In this example, vector addition, scalar multiplication, and additive identity actually coincide with usual addition, usual multiplication, and the number zero, respectively. A linear combination of vectors $v_1,v_2,\ldots,v_n \in \mathbb{R}$ is $$v = \sum\limits_{i=1}^{n} k_i \cdot v_i,$$ i.e., the vector summation coincides with the usual sum.

### A concrete vector space 2

Another example was given by @mrf: If you consider the concrete $\mathbb{R}$-vector space $(\mathbb{R}_{>0}, \cdot, \hat{},1)$, where the symbol $\hat{}$ shall denote usual exponentiation, then vector addition, scalar multiplication, and additive identity correspond to usual multiplication, usual exponentiation, and the number one, respectively. A linear combination of vectors $v_1,v_2,\ldots,v_n \in \mathbb{R}_{>0}$ is $$v = \prod\limits_{i=1}^{n} (v_i)^{k_i},$$ i.e., the vector summation coincides with the usual product.

• ...wow...I seemed to have asked a very experienced question for sitting in elementary linear algebra, thank you for taking the time I will understand your post (wheather it takes me a few years or not i've copy pasted it into my questions folder. Thank you for taking the time Sep 24, 2015 at 19:08