# Scalar multiplication in vector space

I am studying vector spaces and I have problem understanding multiplication of a vector with a scalar from a field over which the vector space is defined.

It is given in a book which I refer that, for every $v$ of a vector space $V$, and $x$ of a field, there is $xv$ in the set $V$.
does it mean number of vectors in a vector space equal the number of elements in a field? and there is an addition satisfying isomorphism?

I think just as the field elements are added, their maps in vectors get added following isomorphism under addition. Am i right?

I think addition of vectors and scalar multiplication work in tandem, but I cant find examples of finite vectors of finite dimensions, to see the addition and meaning of scalar multiplication.

• Does it mean number of vectors in a vector space equal the number of elements in a field? No. You don't necessarily get all the vectors of $V$ starting with a single $x$. Just a 1-dimensional subspace. – Jyrki Lahtonen Sep 8 '17 at 17:06

If $V$ is a vector space over the field $\mathbf F$, then it must satisfy two properties, namely closure under addition and closure under multiplication.
For closure under multiplication, we demand that if $u \in V$, $a \in \mathbf F$, then $a \mathbf F \in V$. Note that the 'multiplication' needs to be defined beforehand.
When you think "vector space", the basic example that you should have in your head is $\mathbb{R}^n$, which is an $n$-dimensional vector space over the field $\mathbb{R}$. If you are studying abstract algebra in the US (or a system that follows a similar general curriculum for teaching mathematics), then you should have some familiarity with $\mathbb{R}^n$ already, either from a multivariable calculus class or a linear algebra class.
A vector in (for example) $\mathbb{R}^3$ can be seen as a $3$-tuple of real numbers, for example $$(1,2,3),\quad (1,0,0),\quad\text{or}\quad (\pi, e, 47).$$ We can multiple vectors in $\mathbb{R}^3$ by a scalar (i.e. real number). In this vector space, the scalar multiplication $c \mathbf{v}$ has the effect of scaling each entry in the tuple. For example, $$4 \cdot (1,2,3) = (4,8,12).$$ One of the axioms of a vector space is that multiplication of a vector by a scalar gives another vector (i.e. the space is closed under scalar multiplication). Observe that $(4,8,12)$ is another $3$-tuple of real numbers, and so lives in $\mathbb{R}^3$.
If you are working over finite fields, you might consider the simplest such example, perhaps $\mathbb{F}_3$ (the field with three elements---$\mathbb{F}_2$ is probably too simple). The $n$-fold Cartesian product of $\mathbb{F}_3$ with itself will form a vector space over $\mathbb{F}_3$, with addition defined coordinatewise, and scalar multiplication working in the same way as in $\mathbb{R}^n$. For example, a typical element of the vector space $\mathbb{F}_3^2$ is a tuple $(a,b)$, where $a,b\in\mathbb{F}_3$. If $(a,b),(c,d)\in\mathbb{F}_3^2$, and $\alpha\in\mathbb{F}_3$ is a scalar, then $$(a,b)+(c,d) = (a+c,b+d) \qquad\text{and}\qquad \alpha(a,b) = (\alpha a, \alpha b).$$ Note that $\mathbb{F}_3$ has three elements, while $\mathbb{F}_3^2$ has nine elements, which appears to give a partial answer to one of your questions.