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Mathematics LibreTexts

3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces

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    19039
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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

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