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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.

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  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Sep 8 '17 at 17:06
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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.

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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.

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  • $\begingroup$ thank you. Does it mean every vector space is just a tuple of some different fields? $\endgroup$ – jnyan Sep 8 '17 at 13:48
  • $\begingroup$ @jnyan Unfortunately, no. A finite dimensional vector space over a field is isomorphic to the Cartesian product, but there are infinite dimensional vector spaces that aren't so nice. My goto example of such objects come from functional analysis, which can be naively thought of as "linear algebra on steroids." $\endgroup$ – Xander Henderson Sep 8 '17 at 13:51

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