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3.1: The Euclidean n-Space, Eⁿ

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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).

Definition

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 $$\overline{x} \| \overline{0}$$.

Example $$\PageIndex{1}$$

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.

Theorem $$\PageIndex{1}$$

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. $$\square$$

Theorem $$\PageIndex{2}$$

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 $$x_{11}, x_{12}, \ldots, x_{1 n}$$.

Theorem $$\PageIndex{3}$$

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 $$a=1, b=0 )$$.

Theorem $$\PageIndex{4}$$

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'). $$square$$

Theorem $$\PageIndex{5}$$

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

$|a||x|=|a x| ;|x+y| \leq|x|+|y| ;|x-y| \geq| | x|-| y| | \quad\left(a, x, y \in E^{1}\right).$