3.1: The Euclidean n-Space, Eⁿ
By definition, the Euclidean \(n\)- space \(E^{n}\) is the set of all possible ordered \(n\)-tuples of real numbers, i.e., the Cartesian product
\[E^{1} \times E^{1} \times \cdots \times E^{1}(n \text{ times}).\]
In particular, \(E^{2}=E^{1} \times E^{1}=\left\{(x, y) | x, y \in E^{1}\right\}\),
\[E^{3}=E^{1} \times E^{1} \times E^{1}=\left\{(x, y, z) | x, y, z \in E^{1}\right\},\]
and so on. \(E^{1}\) itself is a special case of \(E^{n}(n=1).\)In a familiar way, pairs \((x, y)\) can be plotted as points of the \(x y\) -plane, or as "vectors" (directed line segments) joining \((0,0)\) to such points. Therefore, the pairs \((x, y)\) themselves are called points or vectors in \(E^{2} ;\) similarly for \(E^{3}\).
\(\operatorname{In} E^{n}(n>3),\) there is no actual geometric representation, but it is convenient to use geometric language in this case, too. Thus any ordered \(n\)-tuple \(\left(x_{1}, x_{2}, \ldots, x_{n}\right)\) of real numbers will also be called a point or vector in \(E^{n},\) and the single numbers \(x_{1}, x_{2}, \ldots, x_{n}\) are called its coordinates or components. A point in \(E^{n}\) is often denoted by a single letter (preferably with a bar or an arrow above it), and then its \(n\) components are denoted by the same letter, with subscripts (but without the bar or arrow). For example,
\[\overline{x}=\left(x_{1}, \ldots, x_{n}\right), \vec{u}=\left(u_{1}, \ldots, u_{n}\right), etc.;\]
\(\overline{x}=(0,-1,2,4)\) is a point (vector) in \(E^{4}\) with coordinates \(0,-1,2,\) and 4 (in this order). The formula \(\overline{x} \in E^{n}\) means that \(\overline{x}=\left(x_{1}, \ldots, x_{n}\right)\) is a point (vector) in \(E^{n} .\) since such "points" are ordered \(n\) -tuples, \(\overline{x}\) and \(\overline{y}\) are equal \((\overline{x}=\overline{y})\) iff the corresponding coordinates are the same, i.e., \(x_{1}=y_{1}, x_{2}=y_{2}\) \(\ldots, x_{n}=y_{n}\) (see Problem 1 below).
The point whose coordinates are all 0 is called the \(z\) ero-vector or the origin, denoted \(\overrightarrow{0}\) or \(\overline{0} .\) The vector whose \(k\) th component is \(1,\) and the other components are \(0,\) is called the \(k\) th basic unit vector, denoted \(\vec{e}_{k} .\) There are exactly \(n\) such vectors,
\[\vec{e}_{1}=(1,0,0, \ldots, 0), \vec{e}_{2}=(0,1,0, \ldots, 0), \ldots, \vec{e}_{n}=(0, \ldots, 0,1)\]
In \(E^{3},\) we often write \(\overline{i}, \overline{j},\) and \(\vec{k}\) for \(\vec{e}_{1},\) and \((x, y, z)\) for \(\left(x_{1}, x_{2}, x_{3}\right) .\) Similarly in \(E^{2} .\) Single real numbers are called scalars (as opposed to vectors).
Given \(\overline{x}=\left(x_{1}, \ldots, x_{n}\right)\) and \(\overline{y}=\left(y_{1}, \ldots, y_{n}\right)\) in \(E^{n},\) we define the following.
1. The sum of \(\overline{x}\) and \(\overline{y}\),
\[\overline{x}+\overline{y}=\left(x_{1}+y_{1}, x_{2}+y_{2}, \ldots, x_{n}+y_{n}\right) (\text{hence } \overline{x}+\overline{0}=\overline{x} ).\]
2. The dot product, or inner product, of \(\overline{x}\) and \(\overline{y},\)
\[\overline{x} \cdot \overline{y}=x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{n} y_{n}.\]
3. The distance between \(\overline{x}\) and \(\overline{y},\)
\[\rho(\overline{x}, \overline{y})=\sqrt{\left(x_{1}-y_{1}\right)^{2}+\left(x_{2}-y_{2}\right)^{2}+\cdots+\left(x_{n}-y_{n}\right)^{2}}.\]
4. The absolute value, or length, of \(\overline{x},\)
\[|\overline{x}|=\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}=\rho(\overline{x}, \overline{0})=\sqrt{\overline{x} \cdot \overline{x}}\]
(three formulas that are all equal by Definitions 2 and 3\()\).
5. The inverse of \(\overline{x},\)
\[-\overline{x}=\left(-x_{1},-x_{2}, \dots,-x_{n}\right).\]
6. The product of \(\overline{x}\) by a scalar \(c \in E^{1},\)
\[c \overline{x}=\overline{x} c=\left(c x_{1}, c x_{2}, \dots, c x_{n}\right);\]
in particular, \((-1) \overline{x}=\left(-x_{1},-x_{2}, \dots,-x_{n}\right)=-\overline{x}, 1 \overline{x}=\overline{x},\) and \(0 \overline{x}=\overline{0}\).
7. The difference of \(\overline{x}\) and \(\overline{y},\)
\[\overline{x}-\overline{y}=\overrightarrow{y x}=\left(x_{1}-y_{1}, x_{2}-y_{2}, \dots, x_{n}-y_{n}\right).\]
In particular, \(\overline{x}-\overline{0}=\overline{x}\) and \(\overline{0}-\overline{x}=-\overline{x} .\) (Verify!)
Note 1. Definitions \(2-4\) yield scalars , while the rest are vectors .
Note 2. We shall not define inequalities \((<)\) in \(E^{n}(n \geq 2),\) nor shall we define vector products other than the dot product \((2),\) which is a scalar .
Note 3. From Definitions 3, 4, and 7, we obtain \(\rho(\overline{x}, \overline{y})=|\overline{x}-\overline{y}| .\) (Verify!)
Note 4. We often write \(\overline{x} / c\) for \((1 / c) \overline{x},\) where \(c \in E^{1}, c \neq 0\).
Note 5. In \(E^{1}, \overline{x}=\left(x_{1}\right)=x_{1} .\) Thus, by Definition 4,
\[|\overline{x}|=\sqrt{x_{1}^{2}}=\left|x_{1}\right|,\]
where \(\left|x_{1}\right|\) is defined as in Chapter 2, §§1, Definition 4. Thus the two definitions agree .
We call \(\overline{x}\) a unit vector iff its length is \(1,\) i.e., \(|x|=1 .\) Note that if \(\overline{x} \neq \overline{0}\), then \(\overline{x} / | \overline{x} /\) is a unit vector, since
\[\left|\frac{\overline{x}}{|\overline{x}|}\right|=\sqrt{\frac{x_{1}^{2}}{|\overline{x}|^{2}}+\cdots+\frac{x_{n}^{2}}{|\overline{x}|^{2}}}=1.\]
The vectors \(\overline{x}\) and \(\overline{y}\) are said to be orthogonal or perpendicular \((\overline{x} \perp \overline{y})\) iff \(\overline{x} \cdot \overline{y}=0\) and \(\operatorname{parallel}(\overline{x} \| \overline{y})\) iff \(\overline{x}=t \overline{y}\) or \(\overline{y}=t \overline{x}\) for some \(t \in E^{1} .\) Note that \(\overline{x} \perp \overline{0}\) and \(\overline{x} \| \overline{0}\).
If \(\overline{x}=(0,-1,4,2)\) and \(\overline{y}=(2,2,-3,2)\) are vectors in \(E^{4},\) then
\(\begin{aligned} \overline{x}+\overline{y} &=(2,1,1,4); \\ \overline{x}-\overline{y} &=(-2,-3,7,0); \\ \rho(\overline{x}, \overline{y}) &=|\overline{x}-\overline{y}|=\sqrt{2^{2}+3^{2}+7^{2}+0^{2}}=\sqrt{62}; \\(\overline{x}+\overline{y}) \cdot(\overline{x}-\overline{y}) &=2(-2)+1(-3)+7+0=0. \end{aligned}\)
\(\mathrm{So}(\overline{x}+\overline{y}) \perp(\overline{x}-\overline{y})\) here.
For any vectors \(\overline{x}, \overline{y},\) and \(\overline{z} \in E^{n}\) and any \(a, b \in E^{1},\) we have
(a) \(\overline{x}+\overline{y}\) and a \(\overline{x}\) are vectors in \(E^{n}\) (closure laws);
(b) \(\overline{x}+\overline{y}=\overline{y}+\overline{x}\) (commutativity of vector addition);
(c) \((\overline{x}+\overline{y})+\overline{z}=\overline{x}+(\overline{y}+\overline{z})\) (associativity of vector addition);
(d) \(\overline{x}+\overline{0}=\overline{0}+\overline{x}=\overline{x},\) i.e., \(\overline{0}\) is the neutral element of addition;
(e) \(\overline{x}+(-\overline{x})=\overline{0},\) i.e., \(-\overline{x}\) is the additive inverse of \(\overline{x}\);
(f) \(a(\overline{x}+\overline{y})=a \overline{x}+a \overline{y}\) and \((a+b) \overline{x}=a \overline{x}+b \overline{x}\) (distributive laws);
(g) \((a b) \overline{x}=a(b \overline{x})\);
(h) \(1 \overline{x}=\overline{x}\).
- Proof
-
Assertion (a) is immediate from Definitions 1 and \(6 .\) The rest follows from corresponding properties of real numbers.
For example, to prove (b) , let \(\overline{x}=\left(x_{1}, \ldots, x_{n}\right), \overline{y}=\left(y_{1}, \ldots, y_{n}\right) .\) Then by definition, we have
\[\overline{x}+\overline{y}=\left(x_{1}+y_{1}, \dots, x_{n}+y_{n}\right) \text{ and } \overline{y}+\overline{x}=\left(y_{1}+x_{1}, \ldots, y_{n}+x_{n}\right).\]
The right sides in both expressions, however, coincide since addition is commutative \(i n E^{1} .\) Thus \(\overline{x}+\overline{y}=\overline{y}+\overline{x},\) as claimed; similarly for the rest, which we leave to the reader. \(\square\)
If \(\overline{x}=\left(x_{1}, \dots, x_{n}\right)\) is a vector in \(E^{n},\) then, with \(\overline{e}_{k}\) as above,
\[\overline{x}=x_{1} \overline{e}_{1}+x_{2} \overline{e}_{2}+\cdots+x_{n} \overline{e}_{n}=\sum_{k=1}^{n} x_{k} \overline{e}_{k}.\]
Moreover, if \(\overline{x}=\sum_{k=1}^{n} a_{k} \overline{e}_{k}\) for some \(a_{k} \in E^{1},\) then necessarily \(a_{k}=x_{k}\), \(k=1, \ldots, n\).
- Proof
-
By definition,
\[\overline{e}_{1}=(1,0, \ldots, 0), \overline{e}_{2}=(0,1, \ldots, 0), \ldots, \overline{e}_{n}=(0,0, \ldots, 1).\]
Thus
\[x_{1} \overline{e}_{1}=\left(x_{1}, 0, \ldots, 0\right), x_{2} \overline{e}_{2}=\left(0, x_{2}, \dots, 0\right), \ldots, x_{n} \overline{e}_{n}=\left(0,0, \ldots, x_{n}\right).\]
Adding up componentwise, we obtain
\[\sum_{k=1}^{n} x_{k} \overline{e}_{k}=\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\overline{x},\]
as asserted.
Moreover, if the \(x_{k}\) are replaced by any other \(a_{k} \in E^{1},\) the same process yields
\[\left(a_{1}, \ldots, a_{n}\right)=\overline{x}=\left(x_{1}, \ldots, x_{n}\right),\]
i.e., the two \(n\) -tuples coincide, whence \(a_{k}=x_{k}, k=1, \ldots, n\). \(\square\)
Note 6. Any sum of the form
\[\sum_{k=1}^{m} a_{k} \overline{x}_{k} \quad\left(a_{k} \in E^{1}, \overline{x}_{k} \in E^{n}\right)\]
is called a linear combination of the vectors \(\overline{x}_{k}\) (whose number \(m\) is arbitrary). Thus Theorem 2 shows that \(a n y \overline{x} \in E^{n}\) can be expressed, in a unique way, as a linear combination of the \(n\) basic unit vectors. In \(E^{3},\) we write
\[\overline{x}=x_{1} \overline{i}+x_{2} \overline{j}+x_{3} \overline{k}.\]
Note 7. If, as above, some vectors are numbered (e.g., \(\overline{x}_{1}, \overline{x}_{2}, \ldots, \overline{x}_{m} )\), we denote their components by attaching a second subscript; for example, the components of \(\overline{x}_{1}\) are \(x_{11}, x_{12}, \ldots, x_{1 n}\).
For any vectors \(\overline{x}, \overline{y},\) and \(\overline{z} \in E^{n}\) and any \(a, b \in E^{1},\) we have
(a) \(\overline{x} \cdot \overline{x} \geq 0,\) and \(\overline{x} \cdot \overline{x}>0\) iff \(\overline{x} \neq \overline{0}\);
(b) \((a \overline{x}) \cdot(b \overline{y})=(a b)(\overline{x} \cdot \overline{y})\);
(c) \(\overline{x} \cdot \overline{y}=\overline{y} \cdot \overline{x}\) (commutativity of inner products);
(d) \((\overline{x}+\overline{y}) \cdot \overline{z}=\overline{x} \cdot \overline{z}+\overline{y} \cdot \overline{z}(\)distributive \(\operatorname{law})\).
- Proof
-
To prove these properties, express all in terms of the components of \(\overline{x}\), \(\overline{y},\) and \(\overline{z},\) and proceed as in Theorem 1. \(\square\)
Note that (b) implies \(\overline{x} \cdot \overline{0}=0\) (put \(a=1, b=0 )\).
For any vectors \(\overline{x}\) and \(\overline{y} \in E^{n}\) and any \(a \in E^{1},\) we have the following properties:
(a') \(|\overline{x}| \geq 0,\) and \(|\overline{x}|>0\) iff \(\overline{x} \neq \overline{0}\).
(b') \(|a \overline{x}|=|a||\overline{x}|\).
(c') \(|\overline{x} \cdot \overline{y}| \leq|\overline{x}||\overline{y}|,\) or, in components,
\[\left(\sum_{k=1}^{n} x_{k} y_{k}\right)^{2} \leq\left(\sum_{k=1}^{n} x_{k}^{2}\right)\left(\sum_{k=1}^{n} y_{k}^{2}\right) \quad(\text{Cauchy-Schwarz inequality})\]
Equality, \(|\overline{x} \cdot \overline{y}|=|\overline{x}||\overline{y}|,\) holds iff \(\overline{x} \| \overline{y}\).
(d') \(|\overline{x}+\overline{y}| \leq|\overline{x}|+|\overline{y}|\) and \(| | \overline{x}|-| \overline{y}| | \leq|\overline{x}-\overline{y}|\) (triangle inequalities).
- Proof
-
Property (a') follows from Theorem 3\((\mathrm{a})\) since
\[|\overline{x}|^{2}=\overline{x} \cdot \overline{x}( \text{see Definition 4}).\]
For (b'), use Theorem \(3(\mathrm{b}),\) to obtain
\[(a \overline{x}) \cdot(a \overline{x})=a^{2}(\overline{x} \cdot \overline{x})=a^{2}|\overline{x}|^{2}.\]
By Definition \(4,\) however,
\[(a \overline{x}) \cdot(a \overline{x})=|a \overline{x}|^{2}.\]
Thus
\[|a \overline{x}|^{2}=a^{2}|x|^{2}\]
so that \(|a \overline{x}|=|a||\overline{x}|,\) as claimed.
NOW we prove (c'). If \(\overline{x} \| \overline{y}\) then \(\overline{x}=t \overline{y}\) or \(\overline{y}=t \overline{x} ;\) so \(|\overline{x} \cdot \overline{y}|=|\overline{x}||\overline{y}|\) follows by (b'). (Verify!)
Otherwise, \(\overline{x} \neq t \overline{y}\) and \(\overline{y} \neq t \overline{x}\) for all \(t \in E^{1} .\) Then we obtain, for all \(t \in E^{1}\)
\[0 \neq|t \overline{x}-\overline{y}|^{2}=\sum_{k=1}^{n}\left(t x_{k}-y_{k}\right)^{2}=t^{2} \sum_{k=1}^{n} x_{k}^{2}-2 t \sum_{k=1}^{n} x_{k} y_{k}+\sum_{k=1}^{n} y_{k}^{2}.\]
Thus, setting
\[A=\sum_{k=1}^{n} x_{k}^{2}, B=2 \sum_{k=1}^{n} x_{k} y_{k}, \text{ and } C=\sum_{k=1}^{n} y_{k}^{2},\]
we see that the quadratic equation
\[0=A t^{2}-B t+C\]
has \(n o\) real solutions in \(t,\) so its discriminant, \(B^{2}-4 A C,\) must be negative; i.e.,
\[4\left(\sum_{k=1}^{n} x_{k} y_{k}\right)^{2}-4\left(\sum_{k=1}^{n} x_{k}^{2}\right)\left(\sum_{k=1}^{n} y_{k}^{2}\right)<0,\]
proving (c').
To prove (d'), use Definition 2 and Theorem 3(d), to obtain
\[|\overline{x}+\overline{y}|^{2}=(\overline{x}+\overline{y}) \cdot(\overline{x}+\overline{y})=\overline{x} \cdot \overline{x}+\overline{y} \cdot \overline{y}+2 \overline{x} \cdot \overline{y}=|\overline{x}|^{2}+|\overline{y}|^{2}+2 \overline{x} \cdot \overline{y}.\]
But \(\overline{x} \cdot \overline{y} \leq|\overline{x}||\overline{y}|\) by (c'). Thus we have
\[|\overline{x}+\overline{y}|^{2} \leq|\overline{x}|^{2}+|\overline{y}|^{2}+2|\overline{x}||\overline{y}|=(|\overline{x}|+|\overline{y}| |)^{2},\]
whence \(|\overline{x}+\overline{y}| \leq|\overline{x}|+|\overline{y}|,\) as required.
Finally, replacing here \(\overline{x}\) by \(\overline{x}-\overline{y},\) we have
\[|\overline{x}-\overline{y}|+|\overline{y}| \geq|\overline{x}-\overline{y}+\overline{y}|=|\overline{x}|, \text{ or } |\overline{x}-\overline{y}| \geq|\overline{x}|-|\overline{y}|,\]
Similarly, replacing \(\overline{y}\) by \(\overline{y}-\overline{x},\) we get \(|\overline{x}-\overline{y}|-|\overline{y}|-|\overline{x}| .\) Hence
\[|\overline{x}-\overline{y}| \geq \pm(|\overline{x}|-|\overline{y}|),\]
i.e., \(|\overline{x}-\overline{y}| \geq| | \overline{x}|-| \overline{y}| |,\) proving the second formula in (d'). \(square\)
For any points \(\overline{x}, \overline{y},\) and \(\overline{z} \in E^{n},\) we have
(i) \(\rho(\overline{x}, \overline{y}) \geq 0,\) and \(\rho(\overline{x}, \overline{y})=0\) iff \(\overline{x}=\overline{y}\);
(ii) \(\rho(\overline{x}, \overline{y})=\rho(\overline{y}, \overline{x})\);
(iii) \(\rho(\overline{x}, \overline{z}) \leq \rho(\overline{x}, \overline{y})+\rho(\overline{y}, \overline{z})(\)triangle inequality\()\);
- Proof
-
(i) By Definition 3 and Note \(3, \rho(\overline{x}, \overline{y})=|\overline{x}-\overline{y}| ;\) therefore, by Theorem 4\(\left(\mathrm{a}^{\prime}\right)\), \(\rho(\overline{x}, \overline{y})=|\overline{x}-\overline{y}| \geq 0\).
Also, \(|\overline{x}-\overline{y}|>0\) iff \(\overline{x}-\overline{y} \neq 0,\) i.e., iff \(\overline{x} \neq \overline{y} .\) Hence \(\rho(\overline{x}, \overline{y}) \neq 0\) iff \(\overline{x} \neq \overline{y},\) and assertion \((\mathrm{i})\) follows.
(ii) By Theorem \(4\left(\mathrm{b}^{\prime}\right),|\overline{x}-\overline{y}|=|(-1)(\overline{y}-\overline{x})|=|\overline{y}-\overline{x}|,\) so (ii) follows.
(iii) By Theorem 4\(\left(\mathrm{d}^{\prime}\right)\),
\(\rho(\overline{x}, \overline{y})+\rho(\overline{y}, \overline{z})=|\overline{x}-\overline{y}|+|\overline{y}-\overline{z}| \geq|\overline{x}-\overline{y}+\overline{y}-\overline{z}|=\rho(\overline{x}, \overline{z}) . \square\)
Note 8. We also have \(|\rho(\overline{x}, \overline{y})-\rho(\overline{z}, \overline{y})| \leq \rho(\overline{x}, \overline{z}) .\) (Prove it!) The two triangle inequalities have a simple geometric interpretation (which explains their name). If \(\overline{x}, \overline{y},\) and \(\overline{z}\) are treated as the vertices of a triangle, we obtain that the length of a side, \(\rho(\overline{x}, \overline{z})\) never exceeds the sum of the two other sides and is never less their difference.
As \(E^{1}\) is a special case of \(E^{n}\) (in which "vectors" are single numbers), all our theory applies to \(E^{1}\) as well. In particular, distances in \(E^{1}\) are defined by \(\rho(x, y)=|x-y|\) and obey the three laws of Theorem \(5 .\) Dot products in \(E^{1}\) become ordinary products \(x y .\) (Why?) From Theorems \(4\left(\mathrm{b}^{\prime}\right)\left(\mathrm{d}^{\prime}\right),\) we have
\[|a||x|=|a x| ;|x+y| \leq|x|+|y| ;|x-y| \geq| | x|-| y| | \quad\left(a, x, y \in E^{1}\right).\]