4
$\begingroup$

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?

$\endgroup$
8
  • 1
    $\begingroup$ 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). $\endgroup$
    – peek-a-boo
    Commented Aug 31, 2023 at 18:19
  • 5
    $\begingroup$ Are you sure your example is a vector space? ;) $\endgroup$
    – blargoner
    Commented Aug 31, 2023 at 18:47
  • $\begingroup$ 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). $\endgroup$ Commented Aug 31, 2023 at 18:48
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Aug 31, 2023 at 19:18
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Aug 31, 2023 at 20:07

2 Answers 2

6
$\begingroup$

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 $(1,-3,0)$.

$\endgroup$
2
  • $\begingroup$ 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 $\mathbb R.$ $\endgroup$
    – md2perpe
    Commented Sep 1, 2023 at 18:29
  • $\begingroup$ @md2perpe yes it is. $\endgroup$ Commented Sep 1, 2023 at 19:15
6
$\begingroup$

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 $\phi(v) = \ln v.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .