# Example of a vector space with different addition/scalar multiplication operators

I have learned that a vector space is a set of elements called vectors, on which are defined an addition operation and a scalar multiplication operation with the scalars in some field $$F$$. However, all the examples of vector fields I know ($$\mathbb{R}^n$$, $$\mathbb{C}^n$$, $$\mathbb{F}^{(n, n)}$$, etc) all use the "normal" operations of addition and scalar multiplication.

As an example, let's consider the set $$S = \{(x, y) \in \mathbb{R}^2 \lvert x + y = 10\}$$, for instance. With the intuitive definitions of addition and scalar multiplication ($$(a, b) + (c, d) = (a + c, b + d)$$ and $$k(a, b) = (ka, kb)$$), I can verify that this is a vector space, but I can't think of any other definitions of the addition/scalar multiplication operation that still makes this set a vector space.

Could someone provide and explain a nonobvious example of addition/scalar multiplication operations that still keeps the set as a vector space?

• There is a bijection betweem that set and the set $\Bbb{C}^{2023}$. Fix such a bijection $\Phi$. Then, there is a unique structure of a 2023-dimensional complex vector space on the set $S$ such that $\Phi$ becomes an isomorphism (of 2023-dimensional complex vector spaces). Commented Aug 31, 2023 at 18:19
• Are you sure your example is a vector space? ;) Commented Aug 31, 2023 at 18:47
• In practice this sort of thing doesn't really matter. Almost every example of a vector space you'll come across is built out of "normal" addition and scalar multiplication (typically of functions). Commented Aug 31, 2023 at 18:48
• Your set is not a vector space, because it is not closed under the operations: for example, $(10,0)$ lies in your set, but $2(10,0) = (20,0)$ does not. Likewise, even though $(10,0)$ and $(0,10)$ are both elements of $S$, their sum as you have defined them, $(10,0)+(0,10) = (10,10)$ does not lie in $S$. You do not have a vector space. Commented Aug 31, 2023 at 19:18
• By the way, you can make your set $S$ into a vector space (in many different ways), with other "additions" and "scalar multiplications": if $a$ and $b$ are any real numbers with $a+b=10$, you can define addition in $S$ by $(x,y)\oplus(z,w) = (x+z-a,y+w-b)$, and scalar multiplication by $\alpha\odot(x,y) = (\alpha x + (1-\alpha)a,\alpha y + (1-\alpha)b)$, and you will get a vector space. Commented Aug 31, 2023 at 20:07

This is not an obvious example, but you can prove it by straightforward computations.

Consider $$V=\{(x,y,z)\in \mathbb{R}^3: x>0\}$$ with the following operations

$$(x,y,z)\oplus (a,b,c)=(x\cdot a,y+b+3,z+c)$$ $$\alpha \otimes (x,y,z)=(x^\alpha,\alpha \cdot y +3(\alpha -1),\alpha \cdot z)$$

Of course, $$+$$ and $$\cdot$$ are the usual sum and multiplication. It can seem weird, but the null element is $$(1,-3,0)$$.

• This vector space is the direct sum of three vector spaces, $V = V_1 \oplus V_2 \oplus V_3,$ where $V_1$ is the space I present in my answer, $V_2$ is like a translated $\mathbb R,$ and $V_3$ is just $\mathbb R.$ Commented Sep 1, 2023 at 18:29
• @md2perpe yes it is. Commented Sep 1, 2023 at 19:15

An example that I remember from my first year at university:

Take $$V=(0,\infty)$$ with $$v_1\oplus v_2 = v_1 v_2$$ and $$k\otimes v = v^k$$ for $$v,v_1,v_2\in V$$ and $$k\in\mathbb R.$$

This is though isomorphic with $$\mathbb R$$ by the isomorphism $$\phi(v) = \ln v.$$