Coordinates in a Basis

Imagine giving someone directions to a shop. You could say "walk 3 blocks north and 2 blocks east from the town hall" — measuring against the town's street grid. Or you could say "it's 1.2 miles down Main Street, on a bearing of 40°" — measuring against a distance and a compass direction instead. Both descriptions pin down the exact same shop. The shop hasn't moved an inch; only the yardsticks you chose to measure it against have changed.

Vectors work the same way. When you write \begin{bmatrix} 3 \\ 2 \end{bmatrix}, you are quietly using the standard basis: "3 steps along \hat{\imath}, 2 along \hat{\jmath}." That's just one convenient choice of ruler — the (x,y) coordinates you've used your whole life are not some God-given property of the vector; they're the readout of one particular basis. The numbers 3 and 2 are the coordinates of the vector in that basis — the weights of the unique linear combination that reaches it.

Choose a different basis and the very same point gets different coordinates — different numbers describing the identical arrow in space, just as "3 blocks north, 2 east" and "1.2 miles at 40°" describe the identical shop.

If \{\vec{b}_1, \dots, \vec{b}_n\} is a basis for a vector space, then every vector \vec{v} in that space can be written as a linear combination of the basis vectors: \vec{v} = c_1\vec{b}_1 + c_2\vec{b}_2 + \cdots + c_n\vec{b}_n.
Because a basis is linearly independent, this combination is unique — there is exactly one list of weights (c_1, \dots, c_n) that works.
Those weights are the coordinates of \vec{v} with respect to the basis B = \{\vec{b}_1,\dots,\vec{b}_n\}, written [\vec{v}]_B = (c_1, \dots, c_n). Finding them means solving a small system of equations for the weights that rebuild the vector.

Same point, new numbers

The black dot p is fixed in place — it never moves. Tilt the basis vectors \vec{b}_1 and \vec{b}_2 with the slider, and read off how many of each it takes to reach the dot. The coordinates (c_1, c_2) churn as the basis turns, even though the target itself never budges — exactly like the shop from the hook, described by a shifting ruler.

Worked example 1 — the trivial case: the standard basis

Take the vector \vec{v} = \begin{bmatrix} 5 \\ -3 \end{bmatrix}. In the standard basis \{\hat{\imath}, \hat{\jmath}\}, we don't even need to solve anything:

\vec{v} = 5\,\hat{\imath} + (-3)\,\hat{\jmath},

so the coordinates of \vec{v} in the standard basis are (5, -3) — the very same numbers you already wrote down. That sounds too easy to call a "worked example", and that's exactly the point: the standard basis is engineered so reading off coordinates costs zero work. Not every basis is this generous.

Worked example 2 — the real skill: a non-standard basis

Now measure the same vector \vec{v} = (5, -3) against a different basis:

\vec{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \vec{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}.

We need weights c_1, c_2 with c_1 \vec{b}_1 + c_2 \vec{b}_2 = \vec{v}. Writing that out component by component gives a small system:

c_1 + c_2 = 5, \qquad c_1 - c_2 = -3.

Adding the two equations: 2c_1 = 2 \Rightarrow c_1 = 1. Substituting back: 1 + c_2 = 5 \Rightarrow c_2 = 4. Check it: 1\cdot(1,1) + 4\cdot(1,-1) = (1+4,\ 1-4) = (5,-3). ✓

So the coordinates of \vec{v} in basis \{\vec{b}_1, \vec{b}_2\} are (1, 4) — completely different digits from (5, -3), describing the exact same arrow. This is the skill that matters: whenever the basis isn't standard, finding coordinates means solving a linear system for the weights.

Worked example 3 — going backwards: coordinates to vector

The reverse question is just as important. Suppose you're told that a vector has coordinates (2, -1) in the basis \vec{b}_1 = \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \vec{b}_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix} — and you need to know what the vector actually is, in ordinary standard coordinates.

Coordinates are just weights, so rebuild it directly: multiply each basis vector by its weight and add.

\vec{v} = 2\vec{b}_1 + (-1)\vec{b}_2 = 2\begin{bmatrix}2\\0\end{bmatrix} - \begin{bmatrix}1\\3\end{bmatrix} = \begin{bmatrix}4\\0\end{bmatrix} - \begin{bmatrix}1\\3\end{bmatrix} = \begin{bmatrix}3\\-3\end{bmatrix}.

So (2,-1) in that basis is the standard vector (3,-3). Notice there was no system to solve this time — going from coordinates back to the vector is always just a weighted sum, the easy direction. It's finding the coordinates in the first place (Example 2) that takes the work.

A bare pair of numbers like (2, -1) means nothing on its own — it's only coordinates with respect to some basis, and different bases turn it into different vectors. In Worked Example 3, the coordinates (2,-1) named the vector (3,-3) — but in the standard basis, the label (2,-1) would simply be the vector (2,-1). Same digits, two entirely different arrows. Whenever you write coordinates, always say which basis they're measured against.

The flip side is just as important, and easy to get backwards: changing the basis never changes the vector itself — only the numbers describing it. The dot in the interactive figure above never moves as you drag the slider; only its coordinate readout does. It's the same arrow in space the whole time, wearing a different outfit.

Why change basis at all?

Because the right basis can make a hard problem easy. Pick coordinates aligned with the grain of a problem and messy formulas turn simple. This is the secret behind diagonalization and behind principal component analysis, which hunts for the basis that best lines up with the spread of a dataset. Changing basis is just changing your point of view — and a good point of view is half the battle.

Physics students learn this trick early: a block sliding down a frictionless ramp has gravity pulling it straight down, which looks awkward in the standard (x,y) basis — both components of the force are messy trig expressions. Switch to a basis tilted with the ramp — one vector pointing down the slope, one perpendicular to it — and gravity's coordinates suddenly become (mg\sin\theta,\ -mg\cos\theta): one axis carries all the sliding motion, the other just balances against the ramp's surface. Nothing about gravity changed — only the ruler we chose to describe it with. Choosing the basis that matches a problem's natural symmetry is one of the most useful tricks in applied mathematics.

The same idea, industrial-strength, is how image and audio compression works. A photograph stored as ordinary pixel values needs a number for every single pixel. But measured in a cleverly chosen basis (a relative of the ones used in JPEG and MP3), a typical photo's coordinates are mostly tiny numbers close to zero — because most photos are smooth and repetitive, not random static. Throw away the near-zero coordinates and keep only the big ones, and the picture barely changes to the eye, but the file shrinks dramatically. Compression is basis-hunting: find the ruler in which "boring" data has almost no length.

An example: line up with the data

Suppose three points happen to lie on the line y = x: (1,1), (2,2) and (3,3). In the standard basis each needs two numbers, and the bare pairs don't obviously shout "we are collinear".

Now measure against a basis aligned with the line — one vector pointing along it, one across it:

\vec{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \vec{b}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}.

Every point on the line is (k, k) = k\,\vec{b}_1 + 0\,\vec{b}_2, so its coordinates in this basis are simply

(1,1) \to (1, 0), \qquad (2,2) \to (2, 0), \qquad (3,3) \to (3, 0).