Vectors

Arrows and tuples: two faces of one beast

Here's an observation so obvious you'll be annoyed I'm saying it: when you drive to the store, you don't just need to know how far you drove — you need to know which direction you went. Ten kilometers north and ten kilometers south are not the same trip. The thing that captures both magnitude AND direction simultaneously is a vector.

We write a vector in $\mathbb{R}^2$ as a column of components: $\vec{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$

or inline as $\vec{v} = (v_1, v_2)$. The $v_1$ is how far you move horizontally; $v_2$ is how far you move vertically. Draw it as an arrow from some starting point in the direction of those components — that arrow is the vector. The position of the tail doesn't matter; two arrows with the same length and direction are the same vector. This is the crucial thing the Federation of Boring Textbooks glosses over, and I am furious about it every semester: vectors are not stuck to the origin. They're free. A vector is a displacement, not a point. Do not mix them up or I will make you re-read this section.

Vector addition: tip-to-tail

Adding two vectors is stupidly natural. The tip-to-tail rule: put $\vec{w}$'s tail at $\vec{v}$'s tip. The sum $\vec{v} + \vec{w}$ is the arrow from $\vec{v}$'s tail to $\vec{w}$'s tip. Draw it and it's obvious.

Componentwise, it's even simpler: $\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} + \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} v_1 + w_1 \\ v_2 + w_2 \end{pmatrix}$

Add the first components; add the second components. No interaction between components. This works in $\mathbb{R}^n$ for any $n$ — just add all the corresponding components.

The zero vector $\vec{0} = (0, 0)$ is the additive identity: $\vec{v} + \vec{0} = \vec{v}$. It's the arrow with zero length — a point, conceptually. It sits at the origin and does absolutely nothing, which is precisely its job.

Vector subtraction $\vec{v} - \vec{w}$ is the arrow from the tip of $\vec{w}$ to the tip of $\vec{v}$ (when both are drawn from the same base point). Componentwise: subtract the components. File that picture away — it's how you compute "the displacement from $\vec{w}$ to $\vec{v}$."

Drag the vectors below and watch how the sum moves as you manipulate each one:

Scalar multiplication: stretch, shrink, flip

A scalar is just a real number — the word "scalar" is fancy for "something that scales". Multiply a vector by a scalar $c$:

$c \cdot \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} c v_1 \\ c v_2 \end{pmatrix}$

Geometrically: $c > 1$ stretches, $0 < c < 1$ shrinks, $c < 0$ reverses direction, $c = 0$ collapses to $\vec{0}$.

In particular, $-1 \cdot \vec{v} = -\vec{v}$ is the additive inverse: $\vec{v} + (-\vec{v}) = \vec{0}$. This is the "undo" operation for vectors, exactly like negative numbers were the undo for addition (remember the integers? same move).

All the usual arithmetic laws hold: scalar multiplication distributes over vector addition, and vector addition is commutative and associative. These aren't miracles — they're inherited directly from the real-number laws applied componentwise. If you're not sure, try checking $c(\vec{v} + \vec{w}) = c\vec{v} + c\vec{w}$ yourself. It takes twelve seconds and requires zero equipment. Do it. I'll wait.

Length: Pythagoras cashes in

How long is the vector $\vec{v} = (v_1, v_2)$? Draw it as an arrow from the origin. It's the hypotenuse of a right triangle with legs $|v_1|$ and $|v_2|$. The Pythagorean theorem gives us the answer immediately:

$\|\vec{v}\| = \sqrt{v_1^2 + v_2^2}$

This is the norm or magnitude or length of $\vec{v}$ — three words for the same thing, because mathematicians are apparently committed to keeping beginners in a state of low-grade panic. They all mean the same damn thing: how long is the arrow. In $\mathbb{R}^n$, the same formula extends:

$\|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + \cdots + v_n^2}$

A unit vector has length exactly 1. The two flagship unit vectors in $\mathbb{R}^2$ are: $\hat{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad \hat{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

These are the standard basis vectors (file that word away for the next stratum — it'll be very important). Normalizing a vector means dividing by its length to get a unit vector:

$\hat{v} = \frac{\vec{v}}{\|\vec{v}\|}$

This gives you a vector pointing the same direction as $\vec{v}$ but with length 1. Useful whenever you care about direction but not magnitude.

Into higher dimensions

Here's where it gets philosophically interesting. In $\mathbb{R}^3$, a vector is $(v_1, v_2, v_3)$ — three components. You can still picture it as an arrow in 3D space. The length formula is $\sqrt{v_1^2 + v_2^2 + v_3^2}$, and that's still just Pythagoras, applied twice.

In $\mathbb{R}^n$, there's a vector with $n$ components: $(v_1, v_2, \ldots, v_n)$. You cannot picture it as a geometric arrow when $n > 3$. Your brain fails. The pictures fade, the components keep working, and that is the whole fucking point of the algebraic formulation — it doesn't require your visual cortex to cooperate. The machinery works regardless of whether you can visualize it. Welcome to genuine abstraction. It will feel strange. Push through.

This is the first taste of genuine abstraction: we build up intuition in $\mathbb{R}^2$ and $\mathbb{R}^3$, and then the algebra carries us forward where pictures cannot. Get comfortable with that feeling, because it's how the next several strata operate.