# Is it possible to have a different than usual definition of vector addition and scalar multiplication?

Every vector space I've encountered which entries are made up of multiple elements of a certain set $$\mathrm{S}$$, e.g. a coordinate space over $$\mathrm{S}$$, have vector addition and scalar multiplication always defined in the same way. I mean component-wise, respecting the "semantic" (position) of elements, without additional manipulations.

I'd like to ask if this is the only right way to define those operations for these vector spaces. If so, how does this derive from the axioms?

For example if we had defined the sum of two elements of $$\mathbb{R}^{n}$$ in the following way: $$\left[ \begin{array}{c} {x'}_1\\ \vdots\\ {x''}_n \end{array} \right] + \left[ \begin{array}{c} {x''}_1\\ \vdots\\ {x''}_n \end{array} \right] = \left[ \begin{array}{c} 2\left({x'}_1 + {x''}_1\right)\\ \vdots\\ 2\left({x'}_n + {x''}_n\right) \end{array} \right]$$ $$\mathbb{R}^{n}$$ would no longer be a vector space because the vector addition would no longer be associative.

Another example with $$\mathbb{R}^{n}$$. If we had defined the scalar multiplication like this: $$\lambda \left[ \begin{array}{c} {x}_1\\ \vdots\\ {x}_n \end{array} \right] = \left[ \begin{array}{c} \lambda^2 {x}_1\\ \vdots\\ \lambda^2 {x}_n \end{array} \right]$$ we would lost what wikipedia calls the distributivity of scalar multiplication with respect to field addition, i.e. $$(a + b)\mathbf{v} \ne a \mathbf{v} + b\mathbf{v}$$.

It seems to me that the only correct way to define those operations is the usual way, otherwise one or more axioms wouldn't be satisfied, but I am not able to derive it from the axioms.

• This is a nice question. Have you made any progress on proving your conjecture? What have you tried? How far can you get? May 30, 2021 at 18:06
• @saulspatz I am not a real matematician ahah. I do study computer science, I am quite familiar with math of course but this seems at another level of difficulty. Honestly I don't even know from where to start. May 30, 2021 at 18:18
• Definning some "add" and "scalar product" operations so as to achieve a Vector Space is possible in other not usual ways. Same as definning "distance" other than euclidean way. The point is that, most of times, these exotic ways have no application in real world. May 30, 2021 at 19:00
• Every vector space has a basis, so any finite-dimensional vector space (over any field) is isomorphic to the space of $n$-tuples with addition and scalar multiplication defined as usual. Therefore, if we did manage to define a vector space on the $n$-tuples with some different operations, we would not in fact have created anything fundamentally new. I'm not sure if this really answers your question, which is why I've made it a comment. May 30, 2021 at 19:46
• Okay, I'll write an answer. May 30, 2021 at 20:15

An isomorphism in general, is a one-to-one map that preserves the struct of whatever kind of object we're talking about. Since your a computer science student, you're probably familiar with isomorphisms of graphs.

An isomorphism $$\phi$$ of two vector spaces $$V$$ and $$W$$ over the same field $$k$$ is a one-to-one function from $$V$$ onto $$W$$ with the following properties:
$$\phi(v_1+v_2)=\phi(v_1)+\phi(v_2),\ \forall v_1,v_2\in V$$

$$\phi(\lambda v)=\lambda \phi(v),\ \forall v\in V\forall \lambda \in k$$

In the first of these equation $$+$$ on the left-hand side is the addition in $$V$$ and on the right-hand side, it is the addition in $$W$$. Similar remarks apply to the scalar multiplication in the second equation.

By the way, if you don't know what I mean by a field, just think of $$k$$ as the real numbers.

As in graph theory, we consider two isomorphic vector spaces to be the equivalent. They're essentially the same structure, differing only in the names we give the elements. So for $$\mathbb{R}^3$$ we can think of a typical element as $$(a,b,c)$$ or as $$\begin{pmatrix}a\\b\\c\end{pmatrix}$$ or as $$ax^2+bx+c$$ It makes no difference.

Now, any two $$n$$-dimensional vector spaces over a given field are isomorphic. Say that $$V$$ has the basis $$e_1, e_2, \dots e_n$$ and W has the basis $$f_1, f_2, f_n$$. Then the function $$\phi:V\to W$$ defined by $$\phi\left(\sum_{i=1}^na_ie_i\right)=\sum_{i=1}^na_if_i$$ is easily seen to be an isomorphism.

That is, there is one and only one $$n$$-dimensional vector space over $$k$$ and we lose nothing by defining it as the the set of $$n$$-tuples with component-wise addition and scalar multiplication.

Hence my comment that if we did manage to define different operations on the $$n$$-tuples, we wouldn't get anything new.

• Thanks @saulspatz for your detailed answer, now everything is clearer to me! Yes, maybe there are different ways to define the two main operations, but there will always be at least the isomorphism that uses the basis so we won't get any more power. A minor point: in the last formula you are not using the basis e_1, ... ,e_n, f_1, ..., f_n that you have declared just above. May 30, 2021 at 21:03
• @AndreaSimoneCosta Yes, I'm a terrible proofreader. Thank you. May 30, 2021 at 21:11

Perhaps you'd be interested in "Tropical Linear Algebra"? Check out the notion in the first section of Introduction to Tropical Geometry by Antonio LaFace.

• It seems very unlikely that the OP will understand this. May 30, 2021 at 18:50
• Yep, this is definitely too advanced for me. May 30, 2021 at 18:55