Vectors in ℝⁿ: Components, Magnitude and Algebra

$\mathbf{R}^n$ as a set and as a space

Fix a positive integer $n$. As a set, $\mathbf{R}^n$ is the collection of all ordered $n$-tuples of real numbers:

The word ordered carries weight: $(1,2)$ and $(2,1)$ are different elements of $\mathbf{R}^2$. Two elements of $\mathbf{R}^n$ are equal if and only if they agree in every position, so one vector equation is $n$ scalar equations at once. This last remark is used in almost every calculation that follows.

As a space, $\mathbf{R}^2$ is the coordinate plane and $\mathbf{R}^3$ is coordinate three-space. For $n \geq 4$ no picture is available, but every algebraic statement below holds verbatim for all $n$, and that is exactly why the algebra is set up carefully rather than left to the diagram.

Columns, and the transpose

Convention. An element of $\mathbf{R}^n$ is a column:

The real numbers $v_1, \ldots, v_n$ are the components (or entries) of $\mathbf{v}$, and $v_i$ is the $i$-th component. Vectors are set in bold; scalars, which here always means real numbers, are not.

Columns consume vertical space, so running text uses the transpose. For any rectangular array of numbers, the transpose is the array obtained by interchanging its rows with its columns, and is written with a superscript $\mathsf{T}$. For a column $\mathbf{v}$ this gives the single row

Hence $(3,-1,4)^{\mathsf{T}}$ denotes the column of $\mathbf{R}^3$ with entries $3$, $-1$ and $4$, and this is how vectors are written inline from here on. Transposing twice returns the original: $(\mathbf{v}^{\mathsf{T}})^{\mathsf{T}} = \mathbf{v}$.

Points and displacement vectors

The tuple $(3,2)^{\mathsf{T}}$ admits two readings, and both are in constant use.

• As a point: the location in the plane reached from the origin by moving $3$ units in the $x$-direction and $2$ units in the $y$-direction. A point has position and no direction.
• As a displacement vector: the instruction move $3$ units in the $x$-direction and $2$ in the $y$-direction, carried out from wherever one currently is. A displacement has magnitude and direction and no position.

Nothing in the symbols distinguishes the two; the reader supplies the interpretation from context. Because that is an unsatisfactory state of affairs, the convention is fixed now and relied on for the rest of the course.

Convention. A point $P$ of $\mathbf{R}^n$ is identified with its position vector $\mathbf{p}$, the displacement from the origin $O$ to $P$. Under this identification, points and displacement vectors are both elements of $\mathbf{R}^n$ and may be combined by the operations defined below. Where the distinction matters it will be stated in words.

Consequence. If $P$ and $Q$ are points with position vectors $\mathbf{p}$ and $\mathbf{q}$, then the displacement from $P$ to $Q$ is

because travelling from $P$ to $O$ and then from $O$ to $Q$ is the net displacement $-\mathbf{p} + \mathbf{q}$. The order matters: $\overrightarrow{QP} = -\overrightarrow{PQ}$.

The two operations

Definition (addition). For $\mathbf{u}, \mathbf{v} \in \mathbf{R}^n$, the sum $\mathbf{u} + \mathbf{v} \in \mathbf{R}^n$ is defined componentwise: $(\mathbf{u}+\mathbf{v})_i = u_i + v_i$ for $i = 1, \ldots, n$. Addition is defined only between vectors with the same number of components.

Definition (scalar multiplication). For $c \in \mathbf{R}$ and $\mathbf{v} \in \mathbf{R}^n$, the scalar multiple $c\mathbf{v} \in \mathbf{R}^n$ is defined by $(c\mathbf{v})_i = c\,v_i$.

Geometric readings. Addition is tip-to-tail: draw $\mathbf{u}$, then draw $\mathbf{v}$ starting where $\mathbf{u}$ ended; the arrow from the start of $\mathbf{u}$ to the end of $\mathbf{v}$ is $\mathbf{u}+\mathbf{v}$. Drawing both from a common start point instead gives the parallelogram reading: $\mathbf{u}+\mathbf{v}$ is the diagonal from that common start. The two pictures agree, which is a geometric statement of commutativity. Scalar multiplication by $c$ stretches a vector by the factor $|c|$, keeping its direction when $c > 0$ and reversing it when $c < 0$; multiplying by $0$ collapses it to a point.

Two distinguished elements. The zero vector $\mathbf{0} = (0,0,\ldots,0)^{\mathsf{T}}$ satisfies $\mathbf{v} + \mathbf{0} = \mathbf{v}$ for every $\mathbf{v}$; it is the displacement that goes nowhere, and it is the only vector with no direction. The negative of $\mathbf{v}$ is $-\mathbf{v} = (-1)\mathbf{v}$, the vector of the same magnitude pointing the opposite way; it satisfies $\mathbf{v} + (-\mathbf{v}) = \mathbf{0}$.

Definition (subtraction). $\mathbf{u} - \mathbf{v} := \mathbf{u} + (-\mathbf{v})$, so componentwise $(\mathbf{u}-\mathbf{v})_i = u_i - v_i$. Geometrically $\mathbf{u}-\mathbf{v}$ is the displacement from the tip of $\mathbf{v}$ to the tip of $\mathbf{u}$ when both are drawn from a common start: it is the unique vector $\mathbf{x}$ with $\mathbf{v} + \mathbf{x} = \mathbf{u}$.

Worked example. Let $\mathbf{u} = (4,-1,2)^{\mathsf{T}}$ and $\mathbf{v} = (0,3,-5)^{\mathsf{T}}$. Then

Each component was handled separately; no component ever interacts with another under these two operations.

The algebraic laws

The following hold for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbf{R}^n$ and all scalars $c, d \in \mathbf{R}$.

1. $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$ (commutativity of addition)
2. $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$ (associativity of addition)
3. $\mathbf{v} + \mathbf{0} = \mathbf{v}$ (additive identity)
4. $\mathbf{v} + (-\mathbf{v}) = \mathbf{0}$ (additive inverse)
5. $c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}$ (distributivity over a vector sum)
6. $(c + d)\mathbf{v} = c\mathbf{v} + d\mathbf{v}$ (distributivity over a scalar sum)
7. $c(d\mathbf{v}) = (cd)\mathbf{v}$ (compatibility of scalar products)
8. $1\mathbf{v} = \mathbf{v}$ (the scalar $1$ acts trivially)

These are not assumptions. Each is a consequence of the corresponding law for real numbers, applied one component at a time. For law 5, the $i$-th component of the left-hand side is $c(u_i + v_i)$, the $i$-th component of the right-hand side is $c u_i + c v_i$, and these are equal because multiplication distributes over addition in $\mathbf{R}$; since the two vectors agree in every component, they are equal. The remaining seven are established the same way, and a reader who has not done so should verify at least laws 2, 6 and 7 in this manner.

Law 8 looks vacuous but is not: it rules out a scalar multiplication that scales by $c$ and also does something else, and without it the list would not pin the structure down. Note also a derived fact worth recording: $0\mathbf{v} = \mathbf{0}$ for every $\mathbf{v}$, since $0\mathbf{v} = (0+0)\mathbf{v} = 0\mathbf{v} + 0\mathbf{v}$ by law 6, and adding $-(0\mathbf{v})$ to both sides gives $\mathbf{0} = 0\mathbf{v}$. The scalar $0$ on the left and the vector $\mathbf{0}$ on the right are different objects; the notation is standard and the context disambiguates.

Looking ahead. In Section 5 this list is read backwards. Rather than proving the eight laws about $\mathbf{R}^n$, they are imposed as axioms on an arbitrary set with an addition and a scalar multiplication, and any such set is called a vector space. The point of that move is that sets other than $\mathbf{R}^n$ satisfy the same eight laws: polynomials of degree at most $3$, $2 \times 2$ matrices, continuous functions on an interval. Every theorem proved from the eight laws alone then applies to all of them at once. Nothing further is needed about this here; the objects of this section are arrows and tuples.

Magnitude, unit vectors, and the standard notation $\mathbf{e}_i$

In $\mathbf{R}^2$ the vector $\mathbf{v} = (v_1, v_2)^{\mathsf{T}}$ is the hypotenuse of a right triangle with legs $|v_1|$ and $|v_2|$, so Pythagoras gives its length as $\sqrt{v_1^2 + v_2^2}$. In $\mathbf{R}^3$, apply Pythagoras twice: the horizontal shadow of $(v_1,v_2,v_3)^{\mathsf{T}}$ has length $\sqrt{v_1^2+v_2^2}$, and this shadow is perpendicular to the vertical rise $|v_3|$, giving $\sqrt{(\sqrt{v_1^2+v_2^2})^2 + v_3^2} = \sqrt{v_1^2+v_2^2+v_3^2}$.

Definition (magnitude / norm). For $\mathbf{v} \in \mathbf{R}^n$,

For $n \leq 3$ this is a theorem about lengths; for $n \geq 4$ it is the definition of length, chosen because it preserves every property the $n \leq 3$ case enjoys. Note that $\|\mathbf{v}\|$ is a scalar, never a vector, and $\|\mathbf{v}\| \geq 0$ always, with $\|\mathbf{v}\| = 0$ if and only if $\mathbf{v} = \mathbf{0}$.

Proposition (scaling). $\|c\mathbf{v}\| = |c|\,\|\mathbf{v}\|$ for every scalar $c$. Proof. $\|c\mathbf{v}\| = \sqrt{\sum_i (c v_i)^2} = \sqrt{c^2 \sum_i v_i^2} = \sqrt{c^2}\,\sqrt{\sum_i v_i^2} = |c|\,\|\mathbf{v}\|$. The absolute value cannot be dropped: with $c = -3$ and $\|\mathbf{v}\| = 7$ the magnitude is $21$, not $-21$, since a magnitude is never negative. $\square$

Definition (unit vector, normalisation). A vector $\mathbf{u}$ is a unit vector if $\|\mathbf{u}\| = 1$. For $\mathbf{v} \neq \mathbf{0}$, the vector

is a unit vector pointing the same way as $\mathbf{v}$; forming it is called normalising $\mathbf{v}$. It is a unit vector because $\|\hat{\mathbf{v}}\| = \big|1/\|\mathbf{v}\|\big|\,\|\mathbf{v}\| = 1$ by the proposition. The hypothesis $\mathbf{v} \neq \mathbf{0}$ cannot be dropped: $\mathbf{0}$ has no direction and $1/\|\mathbf{0}\|$ is undefined. Consequently the vector of length $L$ in the direction of $\mathbf{v}$ is $L\hat{\mathbf{v}}$.

Example. $\mathbf{v} = (6,-8)^{\mathsf{T}}$ has $\|\mathbf{v}\| = \sqrt{36+64} = 10$, so $\hat{\mathbf{v}} = (3/5, -4/5)^{\mathsf{T}}$, and indeed $(3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 1$.

Notation. Write $\mathbf{e}_i \in \mathbf{R}^n$ for the vector whose $i$-th component is $1$ and whose other components are $0$; thus in $\mathbf{R}^3$, $\mathbf{e}_1 = (1,0,0)^{\mathsf{T}}$, $\mathbf{e}_2 = (0,1,0)^{\mathsf{T}}$, $\mathbf{e}_3 = (0,0,1)^{\mathsf{T}}$. Each $\mathbf{e}_i$ is a unit vector, and $(v_1,\ldots,v_n)^{\mathsf{T}} = v_1\mathbf{e}_1 + \cdots + v_n\mathbf{e}_n$. These are commonly called the standard basis vectors, but basis is a technical term defined in Section 5; until then $\mathbf{e}_i$ is notation only, and no property of a basis is being claimed.

Theorem (triangle inequality). For all $\mathbf{u}, \mathbf{v} \in \mathbf{R}^n$,

with equality if and only if one of $\mathbf{u}, \mathbf{v}$ is a non-negative scalar multiple of the other, that is, when they point the same way (or one is $\mathbf{0}$). A proof is given in Section 3, where the inner product supplies the tool; the geometric content is that going directly is never longer than going by way of an intermediate point.

Concrete instance of strict inequality. Take $\mathbf{u} = (1,0)^{\mathsf{T}}$ and $\mathbf{v} = (0,1)^{\mathsf{T}}$. Then $\mathbf{u}+\mathbf{v} = (1,1)^{\mathsf{T}}$ and $\|\mathbf{u}+\mathbf{v}\| = \sqrt{2} \approx 1.414$, whereas $\|\mathbf{u}\| + \|\mathbf{v}\| = 2$. So $\|\mathbf{u}+\mathbf{v}\|$ and $\|\mathbf{u}\|+\|\mathbf{v}\|$ are genuinely different quantities. Instance of equality. Take $\mathbf{u} = (1,0)^{\mathsf{T}}$ and $\mathbf{v} = (3,0)^{\mathsf{T}}$: then $\|\mathbf{u}+\mathbf{v}\| = 4 = 1 + 3$.

Corollary (reverse triangle inequality). $\|\mathbf{u}+\mathbf{v}\| \geq \big|\,\|\mathbf{u}\| - \|\mathbf{v}\|\,\big|$. Proof. $\|\mathbf{u}\| = \|(\mathbf{u}+\mathbf{v}) + (-\mathbf{v})\| \leq \|\mathbf{u}+\mathbf{v}\| + \|\mathbf{v}\|$, so $\|\mathbf{u}\| - \|\mathbf{v}\| \leq \|\mathbf{u}+\mathbf{v}\|$; interchanging $\mathbf{u}$ and $\mathbf{v}$ gives $\|\mathbf{v}\| - \|\mathbf{u}\| \leq \|\mathbf{u}+\mathbf{v}\|$, and together these bound the absolute value. $\square$ So for $\|\mathbf{u}\| = 5$ and $\|\mathbf{v}\| = 9$, the only possible values of $\|\mathbf{u}+\mathbf{v}\|$ are those in $[4, 14]$.

Distance. The distance between points $P$ and $Q$ with position vectors $\mathbf{p}, \mathbf{q}$ is $\|\mathbf{q}-\mathbf{p}\|$, the magnitude of the displacement between them. This is the same formula as the coordinate distance formula, restated in vector language.