
# 3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces

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I. We shall now follow the pattern of $$E^{n}$$ to obtain the general notion of a vector space (just as we generalized $$E^{1}$$ to define fields).

Let $$V$$ be a set of arbitrary elements (not necessarily $$n$$ -tuples), called "vectors" or "points," with a certain operation (call it "addition," $$+)$$ somehow defined in $$V .$$ Let $$F$$ be any field (e.g. $$, E^{1}$$ or $$C$$); its elements will be called scalars; its zero and unity will be denoted by 0 and $$1,$$ respectively. Suppose that yet another operation ("multiplication of scalars by vectors") has been defined that assigns to every scalar $$c \in F$$ and every vector $$x \in V$$ a certain vector, denoted $$c x$$ or $$x c$$ and called the $$c$$ -multiple of $$x .$$ Furthermore, suppose that this multiplication and addition in $$V$$ satisfy the nine laws specified in Theorem 1 of §§1-3. That is, we have closure:

$(\forall x, y \in V)(\forall c \in F) \quad x+y \in V \text{ and } c x \in V$

Vector addition is commutative and associative. There is a unique zero-vector, $$\overrightarrow{0},$$ such that

$(\forall x \in V) \quad x+\overrightarrow{0}=x$

and each $$x \in V$$ has a unique inverse, $$-x,$$ such that

$x+(-x)=\overrightarrow{0}.$

We have distributivity:

$a(x+y)=a x+a y \text{ and } (a+b) x=a x+b x.$

Finally, we have

$1 x=x$

and

$(a b) x=a(b x)$

$$(a, b \in F ; x, y \in V).$$

In this case, $$V$$ together with these two operations is called a vector space (or a linear space) over the field $$F ; F$$ is called its scalar field, and elements of $$F$$ are called the scalars of $$V$$.

Example $$\PageIndex{1}$$

(a) $$E^{n}$$ is a vector space over $$E^{1}$$ (its scalar field).

(a') $$R^{n},$$ the set of all rational points of $$E^{n}$$ (i.e., points with rational coordinates is a vector space over $$R,$$ the rationals in $$E^{1} .$$ (Note that we could take $$R$$ as a scalar field for all of $$E^{n} ;$$ this would yield another vector space, $$E^{n}$$ over $$R,$$ not to be confused with $$E^{n}$$ over $$E^{1},$$ i.e., the ordinary $$E^{n} . )$$

(b) Let $$F$$ be any field, and let $$F^{n}$$ be the set of all ordered $$n$$ -tuples of elements of $$F,$$ with sums and scalar multiples defined as in $$E^{n}$$ (with $$F$$ playing the role of $$E^{1} ) .$$ Then $$F^{n}$$ is a vector space over $$F($$ proof as in Theorem 1 of §§1-3).

(c) Each field $$F$$ is a vector space (over itself) under the addition and multiplication defined in $$F .$$ Verify!

(d) Let $$V$$ be a vector space over a field $$F,$$ and let $$W$$ be the set of all possible mappings

$f : A \rightarrow V$

from some arbitrary set $$A \neq \emptyset$$ into $$V .$$ Define the sum $$f+g$$ of two such maps by setting

$(f+g)(x)=f(x)+g(x) \text{ for all } x \in A.$

Similarly, given $$a \in F$$ and $$f \in W,$$ define the map $$a f$$ by

$(a f)(x)=a f(x).$

Vector spaces over $$E^{1}$$ (respectively, $$C )$$ are called real (respectively, complex) linear spaces. Complex spaces can always be transformed into real ones by restricting their scalar field $$C$$ to $$E^{1}$$ (treated as a subfield of $$C )$$.

II. An important example of a complex linear space is $$C^{n},$$ the set of all ordered $$n$$-tuples

$x=\left(x_{1}, \ldots, x_{n}\right)$

of complex numbers $$x_{k}$$ (now treated as scalars), with sums and scalar multiples defined as in $$E^{n} .$$ In order to avoid confusion with conjugates of complex numbers, we shall not use the bar notation $$\overline{x}$$ for a vector in this section, writing simply $$x$$ for it. Dot products in $$C^{n}$$ are defined by

$x \cdot y=\sum_{k=1}^{n} x_{k} \overline{y}_{k},$

where $$\overline{y}_{k}$$ is the conjugate of the complex number $$y_{k}$$ (see §8), and hence a scalar in $$C .$$ Note that $$\overline{y}_{k}=y_{k}$$ if $$y_{k} \in E^{1}$$ . Thus, for vectors with real components,

$x \cdot y=\sum_{k=1}^{n} x_{k} y_{k},$

as in $$E^{n} .$$ The reader will easily verify (exactly as for $$E^{n}$$) that, for $$x, y \in C^{n}$$ and $$a, b \in C,$$ we have the following properties:

1. $$x \cdot y \in C ;$$ thus $$x \cdot y$$ is a scalar, not a vector.

2. $$x \cdot x \in E^{1},$$ and $$x \cdot x \geq 0 ;$$ moreover, $$x \cdot x=0$$ iff $$x=\overrightarrow{0} .$$ (Thus the dot product of a vector by itself is a real number $$\geq 0 . )$$

3. $$x \cdot y=\overline{y \cdot x}(=$$ conjugate of $$y \cdot x) .$$ Commutativity fails in general.

4. $$(a x) \cdot(b y)=(a \overline{b})(x \cdot y) .$$ Hence $$\left(\mathrm{iv}^{\prime}\right)(a x) \cdot y=a(x \cdot y)=x \cdot(\overline{a} y)$$.

5. $$(x+y) \cdot z=x \cdot z+y \cdot z$$ and $$\left(\mathrm{5}^{\prime}\right) z \cdot(x+y)=z \cdot x+z \cdot y$$.

Observe that (5') follows from (5) by (3). (Verify!)

III. Sometimes (but not always) dot products can also be defined in real or complex linear spaces other than $$E^{n}$$ or $$C^{n},$$ in such a manner as to satisfy the laws (1)-(5), hence also (5'), listed above, with $$C$$ replaced by $$E^{1}$$ if the space is real. If these laws hold, the space is called Euclidean. For example, $$E^{n}$$ is a real Euclidean space and $$C^{n}$$ is a complex one.

In every such space, we define absolute values of vectors by

$|x|=\sqrt{x \cdot x}.$

(This root exists in $$E^{1}$$ by formula (ii).) In particular, this applies to $$E^{n}$$ and $$C^{n} .$$ Then given any vectors $$x, y$$ and a scalar $$a,$$ we obtain as before the following properties:

(a') $$|x| \geq 0 ;$$ and $$|x|=0$$ iff $$x=\overrightarrow{0}$$.

(b') $$|a x|=|a||x|$$.

(c') Triangle inequality: $$|x+y| \leq|x|+|y|$$.

(d') Cauchy-Schwarz inequality: $$|x \cdot y| \leq|x||y|,$$ and $$|x \cdot y|=|x||y|$$ iff $$x \| y$$ (i.e., $$x=a y$$ or $$y=a x$$ for some scalar $$a ) .$$

We prove only (d') ;\) the rest is proved as in Theorem 4 of §§1-3.

If $$x \cdot y=0,$$ all is trivial, so let $$z=x \cdot y=r c \neq 0,$$ where $$r=|x \cdot y|$$ and $$c$$ has modulus $$1,$$ and let $$y^{\prime}=c y .$$ For any (variable $$) t \in E^{1},$$ consider $$\left|t x+y^{\prime}\right| .$$ By definition and (5),(3), and (4),

\begin{aligned}\left|t x+y^{\prime}\right|^{2} &=\left(t x+y^{\prime}\right) \cdot\left(t x+y^{\prime}\right) \\ &=t x \cdot t x+y^{\prime} \cdot t x+t x \cdot y^{\prime}+y^{\prime} \cdot y^{\prime} \\ &=t^{2}(x \cdot x)+t\left(y^{\prime} \cdot x\right)+t\left(x \cdot y^{\prime}\right)+\left(y^{\prime} \cdot y^{\prime}\right) \end{aligned}

since $$\overline{t}=t .$$ Now, since $$c \overline{c}=1$$,

$x \cdot y^{\prime}=x \cdot(c y)=(\overline{c} x) \cdot y=\overline{c} r c=r=|x \cdot y|.$

Similarly,

$y^{\prime} \cdot x=\overline{x \cdot y^{\prime}}=\overline{r}=r=|x \cdot y|, x \cdot x=|x|^{2}, \text{ and } y^{\prime} \cdot y^{\prime}=y \cdot y=|y|^{2}.$

Thus we obtain

$\left(\forall t \in E^{1}\right) \quad|t x+c y|^{2}=t^{2}|x|^{2}+2 t|x \cdot y|+|y|^{2}.$

Here $$|x|^{2}, 2|x \cdot y|,$$ and $$|y|^{2}$$ are fixed real numbers. We treat them as coefficients in $$t$$ of the quadratic trinomial

$f(t)=t^{2}|x|^{2}+2 t|x \cdot y|+|y|^{2}.$

Now if $$x$$ and $$y$$ are not parallel, then $$c y \neq-t x,$$ and so

$|t x+c y|=\left|t x+y^{\prime}\right| \neq 0$

for any $$t \in E^{1} .$$ Thus by $$(1),$$ the quadratic trinomial has no real roots; hence its discriminant,

$4|x \cdot y|^{2}-4(|x||y|)^{2},$

is negative, so that $$|x \cdot y|<|x||y|.$$

If, however, $$x \| y,$$ one easily obtains $$|x \cdot y|=|x||y|,$$ by $$\left(\mathrm{b}^{\prime}\right) .$$ (Verify.)

Thus $$|x \cdot y|=|x||y|$$ or $$|x \cdot y|<|x||y|$$ according to whether $$x \| y$$ or not. $$\square$$

In any Euclidean space, we define distances by $$\rho(x, y)=|x-y| .$$ Planes, lines, and line segments are defined exactly as in $$E^{n} .$$ Thus

$\text{line } \overline{p q}=\left\{p+t(q-p) | t \in E^{1}\right\}( \text{in real and complex spaces alike}).$