# Properties of vector spaces.

I am just learning about vector spaces and have been given the following definition

"Let V be a set of elements $\textbf{x,y,u,v}$ etc.

And K is a field consisting of elements called scalars

And we define the rules

1. Addition: a binary operation denoted +. To and $\textbf{x,y}\in V$ this rule asigns an element $\textbf{z}\in V :\textbf{z=x+y}$

2. Scalar multiplication: To any $a\in K$ and $\textbf{x}\in V$ this rule assigns an element $\textbf{z}\in V: \textbf{z}=a\textbf{x}$

then V is a vector space over K and the elements of V are called vectors if the following axioms hold..."

and the standard 8 axioms are given.

I have a couple of questions regarding this:

1. Am I correct in thinking that the elements of V, i.e. the 'vectors', may or may not be sets themselves, as ordinary vectirs are?

2. Secondly, is V necessaily an infinite set except for the case when the only element of K is 0? Otherwise the addition and/or scalar multiplication rules would allow you to keep churning out more 'vectors' which are defined also to be elements of V?

• Could the down-voter please let me know what is wrong with the question I have asked? – Meep Apr 27 '17 at 14:28
• Enumerating the elements of $V$ in a list implies that $V$ is countable, which it may not be. Yes, the elements of $V$ might themselves be sets. Any finite-dimensional vector space with a finite scalar field $K$ will have finitely many elements. – Myridium Apr 27 '17 at 14:30
• Don't confuse, e.g., a vector, an $n$-tuple of $\mathbb R^n$, as being a set. Rather, any unique n-tuple $\in R^n$ is simply an element of $\mathbb R^n$, and is not itself a set. For example, $\langle 1, \sqrt 2, \frac 32, 3\rangle \in R^4 = \{\langle x_1, x_2, x_3, x_4\rangle\mid x_i \in \mathbb R,1 \leq i\leq 4\}$. But the vector $\langle 1, \sqrt 2, \frac 32, 3\rangle$ is not a set. I say this because you ask about 'vectors', which may or may not be sets themselves, as "ordinary" vectors are. – amWhy Apr 27 '17 at 16:34

## 1 Answer

(1). The vectors may be sets, or may not. It depends if $V$ is a set of sets or not. All we can say is they are elements of $V.$ However, if by "normal vectors" you mean vectors in ${R}^n$, these are not sets. They are n-tuples.

(2). $V$ need not be infinite. Take the example $k={1}$, $\;V=\{0,1\}$ with $1+1=0.$

• I simply polished things off with formatting. Please feel free to edit whatever you feel you need to edit, if I got anything wrong. You might also want to take a look at the edit review by clicking on the link above the icon of the most recent edit ("edited _ min ago") above my recent edit, to see, e.g., to post {0, 1} you need to write \{0, 1\} – amWhy Apr 27 '17 at 16:45
• Thank you for your reply! As to throw first point, are n-tuples not ordered sets? This is what we were told in our lecturer :/ Also, I am guessing that in the second point 1 and 0 are placements for objects that we call vectors, and not the numbers 1 and 0 themselves since 1+1\neq 0? I.e. we define the sum of this vector object, whatever it is, to be 0? – Meep Apr 27 '17 at 19:10
• Sorry as to the second point again, could we not then add another 1 onto 0 and get -1? Which is not in V? – Meep Apr 27 '17 at 19:14
• @21joanna12 sorry for the slow response. (1). No, n-tuples are not ordered sets. $(3,3)$ is not an "ordered set" containing $3$, it is its own object distinct from the scalar $3$. By definition, a set cannot be ordered, it is merely a collection of mathematical objects. (2). No. $0+1=1+0=1$. Remember that $-x$ is defined as the vector such that $x+(-x)=0$, so in this case $-1=1$. – Aperson123 Apr 28 '17 at 15:52
• @Aperson123 thank you for your reply and your patience with my questions. So in this case 1 is its own inverse. And this does not necessarily mean that 1=0 from the axioms? – Meep Apr 28 '17 at 16:48