Scaling a Vector

Push a plane's throttle forward and its velocity vector doesn't change direction — it's still flying the same way — it just gets longer. Double the throttle, double the speed, and the velocity vector stretches to twice its length while pointing exactly where it pointed before. Pull the throttle into reverse thrust, though, and something more dramatic happens: the vector doesn't just shrink, it flips right around to point the opposite way.

Both of these — stretching and flipping — are the same operation: scalar multiplication, multiplying a vector by an ordinary number (a scalar). It's the simplest way to change a vector, and it never tilts the vector off the line it already lies on.

In components, multiply every entry by the scalar:

c\begin{bmatrix} v_x \\ v_y \end{bmatrix} = \begin{bmatrix} c\,v_x \\ c\,v_y \end{bmatrix}.

So 3\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \end{bmatrix}, and -1\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -2 \\ -1 \end{bmatrix}.

Stretch, shrink, flip

Drag the scalar c below. The faint arrow is the original \vec{v}; the bold one is c\,\vec{v}. Watch how it always stays on the same line through the origin — scaling can lengthen, shorten or reverse a vector, but never tilt it. Slide c from a big positive number down through 1, through 0, and out into negative numbers, and watch the bold arrow shrink to nothing at c = 0 before springing out the other side of the origin.

Worked example: stretching

Let \vec{v} = (2, -1) and scale it by c = 3. Multiply each component:

3(2, -1) = (3 \cdot 2,\ 3 \cdot (-1)) = (6, -3).

The new vector (6, -3) points in exactly the same direction as (2, -1) — check that one is just 3 times the other, component for component — but it's three times as long.

Worked example: flipping with a negative scalar

Now take \vec{v} = (1, 3) and scale it by c = -2:

-2(1, 3) = (-2 \cdot 1,\ -2 \cdot 3) = (-2, -6).

Two things happened at once: the length doubled (because |-2| = 2), and the direction reversed completely (because the scalar was negative). If (1,3) points up and to the right, (-2,-6) points down and to the left — the exact opposite way, just twice as far from the origin.

Worked example: shrinking towards length one

Scalars smaller than 1 shrink a vector instead of stretching it. Take \vec{v} = (3, 4), which happens to have length 5 (from the right-triangle sides 3 and 4). Scale it by c = \tfrac{1}{5}:

\tfrac{1}{5}(3, 4) = (0.6,\ 0.8).

That new vector points exactly where (3,4) did, but its length has shrunk all the way down to 1. Scaling any non-zero vector by one over its own length always produces a length-one vector pointing the same way — the trick behind unit vectors, which you'll meet properly very soon.

The words for it

All the multiples of a single non-zero vector — every c\,\vec{v} as c ranges over all numbers — trace out an entire line through the origin. Scaling by 0 gives the zero vector; scaling by 1 leaves the vector unchanged. Two vectors that are scalar multiples of each other are called parallel (or collinear) — a fact we'll lean on when we meet linear independence.

For any vectors \vec{u}, \vec{v} and scalars c, d:

Associative: c(d\,\vec{v}) = (cd)\,\vec{v} — scaling twice in a row is the same as scaling once by the combined number.
Distributive over vector addition: c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v} — you can scale before or after adding, same result.
Distributive over scalar addition: (c + d)\,\vec{v} = c\vec{v} + d\vec{v} — combining scalars first or scaling separately and adding both work.

Worked example: checking a property

Let's check the associative rule with numbers. Take \vec{v} = (1, 2), c = 2 and d = 3. Scaling twice in a row:

2\big(3(1,2)\big) = 2(3, 6) = (6, 12).

And combining the scalars first, cd = 6, then scaling once:

6(1, 2) = (6, 12).

Same answer either way — exactly what the theorem promises, and a handy thing to know: you never need to apply several scalings in sequence when one combined scaling will do.

Working backwards: finding the scalar

Sometimes you're handed both vectors and asked what scalar connects them. Suppose \vec{v} = (2, 3) and you're told c\,\vec{v} = (6, 9) for some number c. Since every component scales by the same c, check either one:

c = \frac{6}{2} = 3 \qquad \text{and, as a check,} \qquad c = \frac{9}{3} = 3.

Both components agree on c = 3, so (6,9) really is a scalar multiple of (2,3) — the two vectors are parallel. If the two components had given different values of c (say 3 from one and 4 from the other), that would tell you straight away that the second vector is not a scalar multiple of the first — it points off in some other direction entirely.

How length changes when you scale

Scaling a vector by c multiplies its length by |c| (the size of c, ignoring any minus sign):

|c\,\vec{v}| = |c|\cdot|\vec{v}|.

Check it against the very first worked example: \vec{v} = (2,-1) has length \sqrt{2^2 + (-1)^2} = \sqrt{5}. Scale it by c = 3 to get (6,-3), whose length is \sqrt{6^2 + (-3)^2} = \sqrt{45} = 3\sqrt{5} — exactly 3 times as long, just as the rule predicts. Notice the absolute value: whether c is 3 or -3, the length triples either way — it's only the direction that cares about the sign. You'll put this exact formula to work when you meet the magnitude of a vector properly.

We have William Rowan Hamilton to thank for the word itself. While building his 1840s theory of quaternions — four-part numbers meant to capture both a direction and a stretch factor in one object — Hamilton needed a name for the "plain number" part that carried no direction at all, only size. He borrowed from the Latin scala, meaning "ladder" or "scale" (as in a measuring scale), because such a number simply told you how far up or down the scale to slide a quantity — precisely the stretch-shrink-flip job scalar multiplication still does today.

Every pixel's colour in a digital photo is stored as a little vector of three numbers — (\text{red}, \text{green}, \text{blue}). When you drag a "brightness" slider up in a photo editor, the software takes every one of those colour vectors and multiplies it by the same scalar, something like 1.2. Every channel grows by the same factor, the colours stay in the same ratio to each other (so a pixel doesn't change hue, just intensity), and the whole photo looks uniformly brighter. Slide it below 1 and every colour vector shrinks towards black instead.

Mistake one: scaling only part of the vector. A scalar multiplies every component, not just the first one. If you scale (4, 5) by 2 and write down (8, 5) — doubling the x but forgetting the y — the result no longer points the same way at all. The correct answer touches both numbers: (8, 10).

Mistake two: thinking a negative scalar just "makes it negative." Multiplying by -1 doesn't merely flip the signs of the numbers in some abstract sense — it rotates the arrow 180° so it points the exact opposite way in space. A negative scalar always reverses direction; it's only the size of |c| that controls whether the result is longer or shorter.

Open a map app and pinch to zoom in on a landmark. Behind the scenes, every point on screen is stored as a position vector measured from some centre, and "zoom 2×" literally means: multiply every one of those vectors by the scalar 2. Points near the centre barely move; points far from the centre — whose vectors are longer — shoot outward much further, which is exactly why zooming makes the whole picture spread out from the middle rather than sliding sideways.

Game engines and design software use the identical trick to resize objects: scale every vertex's position vector (relative to the object's centre) by the same number, and the whole shape grows or shrinks uniformly without changing its form.