# 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

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