Vector space/Introduction/Section

The addition of two arrows $a$ and $b$, a typical example for vectors.

The central concept of linear algebra is a vector space.

Definition

Let $K$ denote a field, and $V$ a set with a distinguished element $0\in V$, and with two mappings

$+:V\times V⟶V,(u,v)⟼u+v,$

and

$\cdot :K\times V⟶V,(s,v)⟼sv=s\cdot v.$

Then $V$ is called a $K$-vector space (or a vector space over $K$), if the following axioms hold (where $r,s\in K$ and $u,v,w\in V$ are arbitrary).

1. $u+v=v+u$,
2. $(u+v)+w=u+(v+w)$,
3. $v+0=v$,
4. For every $v$, there exists a $z$ such that $v+z=0$,
5. $1\cdot u=u$,
6. $r(su)=(rs)u$,
7. $r(u+v)=ru+rv$,
8. $(r+s)u=ru+su$.

The binary operation in $V$ is called (vector-)addition, and the operation $K\times V\to V$ is called scalar multiplication. The elements in a vector space are called vectors, and the elements $r\in K$ are called scalars. The null element $0\in V$ is called null vector, and for $v\in V$, the inverse element, with respect to the addition, is called the negative of $v$, denoted by $-v$.

The field that occurs in the definition of a vector space is called the base field. All the concepts of linear algebra refer to such a base field. In case $K=ℝ$, we talk about a real vector space, and in case $K=ℂ$, we talk about a complex vector space. For real and complex vector spaces, there exist further structures like length, angle, inner product. But first we develop the algebraic theory of vector spaces over an arbitrary field.

Example

Let $K$ denote a field, and let $n\in {\mathbb{N}}_{+}$. Then the product set

$K^n=\underset{n\text{-times}}{\underbrace{K\times \cdots \times K}}=\{(x_1,\dots ,x_n)\mid x_i\in K\},$

with componentwise addition and with scalar multiplication given by

$s(x_1,\dots ,x_n)=(s{x}_1,\dots ,s{x}_n),$

is a vector space. This space is called the $n$-dimensional standard space. In particular, $K^1=K$ is a vector space.

The null space $0$, consisting of just one element $0$, is a vector space. It might be considered as $K^0=0$.

The vectors in the standard space $K^n$ can be written as row vectors

$(a_1,a_2,\dots ,a_n)$

or as column vectors

$\left(\begin{array}{c}{a}_1\\ a_2\\ ⋮\\ a_n\end{array}\right).$

The vector

$e_i:=\left(\begin{array}{c}0\\ ⋮\\ 0\\ 1\\ 0\\ ⋮\\ 0\end{array}\right),$

where the $1$ is at the $i$-th position, is called $i$-th standard vector.

Example

The complex numbers ${\displaystyle }$ form a field, and therefore they form also a vector space over the field $ℂ$ itself. However, the set of complex numbers equals ${ℝ}^2$ as an additive group. The multiplication of a complex number $a+b\mathrm{i}$ with a real number $s=(s,0)$ is componentwise, so this multiplication coincides with the scalar multiplication on ${ℝ}^2$. Hence, the set of complex numbers is also a real vector space.

Example

For a field $K$, and given natural numbers $m,n$, the set

${\mathrm{Mat}}_{m\times n}(K)$

of all $m\times n$-matrices, endowed with componentwise addition and componentwise scalar multiplication, is a $K$-vector space. The null element in this vector space is the null matrix

$0=\left(\begin{array}{ccc}0& \dots & 0\\ ⋮& \ddots & ⋮\\ 0& \dots & 0\end{array}\right).$

Example

Let $R=K[X]$ be the polynomial ring in one variable over the field $K$, consisting of all polynomials, that is, expressions of the form

$a_n{X}^n+a_{n-1}{X}^{n-1}+\cdots +a_2{X}^2+a_{1}X+a_0,$

with $a_i\in K$. Using componentwise addition and componentwise multiplication with a scalar $s\in K$ (this is also multiplication with the constant polynomial $s$), the polynomial ring is a $K$-vector space.

Example

We consider the inclusion $\mathbb{Q}\subseteq ℝ$ of the rational numbers inside the real numbers. Using the real addition and the multiplication of rational numbers with real numbers, we see that $ℝ$ is a $\mathbb{Q}$-vector space, as follows directly from the field axioms. This is a quite crazy vector space.

Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space. Then the following properties hold (for

$v\in V$ and $s\in K$).

1. We have $0v=0$.
2. We have $s0=0$.
3. We have $(-1)v=-v$.
4. If $s\ne 0$ and $v\ne 0$, then $sv\ne 0$.

Proof

See exercise.

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