Coordinate Systems

Main idea

Once you pick a basis $\mathcal B$ , every vector has exactly one set of coordinates $[\vec x]_{\mathcal B}$ : the weights that build it from the basis vectors. Change the basis and the coordinates change, even though the vector itself doesn't move.

\vec x = c_1\vec b_1 + \cdots + c_n\vec b_n, \qquad [\vec x]_{\mathcal B} = \begin{bmatrix}c_1\\ \vdots\\ c_n\end{bmatrix}, \qquad \vec x = {\htmlData{sym=0}{P_{\mathcal B}}}[\vec x]_{\mathcal B}

What the symbols mean

$P_{\mathcal B}$
Change-of-coordinates matrix
Its columns are the basis vectors.

Explanation

Coordinates relative to a basis are simply the weights in the recipe that builds a vector out of the basis vectors. They're the same "coordinates" you know from the $xy$ -plane: the point $(3, 2)$ has always meant $3$ of $\vec e_1$ plus $2$ of $\vec e_2$ . All that changes now is that you get to pick other building blocks. That's useful because it turns any vector space, even polynomials, into plain lists of numbers in $\mathbb R^n$ . All we're really saying in the formula is: $\vec x = c_1\vec b_1 + \cdots + c_n\vec b_n$ (build $\vec x$ from the basis), $[\vec x]_{\mathcal B}$ (the list of weights, stacked in a column) and $\vec x = P_{\mathcal B}[\vec x]_{\mathcal B}$ (multiplying the basis matrix by the weights rebuilds $\vec x$ , because a matrix times a vector is a combination of its columns). It's three ways of saying the same thing; bear with it.

Uniqueness is where independence from last lesson pays off. If a vector had two sets of weights, subtracting one recipe from the other would give a nontrivial combination of the basis vectors equal to zero, and independence says that can't happen.

Converting: from coordinates to the vector, multiply by $P_{\mathcal B}$ ; from the vector to coordinates, solve $P_{\mathcal B}\vec c = \vec x$ (or use $P_{\mathcal B}^{-1}$ ).

Abstract spaces become $\mathbb R^n$ . In $\mathbb P_2$ with basis $\{1, t, t^2\}$ , the polynomial $3 + 2t - t^2$ has coordinates $(3, 2, -1)$ . Questions about polynomials turn into questions about vectors, and you already know how to answer those.

A common mistake. It's tempting to think a point's coordinates belong to the point, because in the standard basis the coordinates and the point look identical. They don't: the same point has different coordinates in a different basis. There's an animation further down that shows one fixed point getting new coordinates as the basis swings around.

See it

The point x = (2, 3) never moves here. Only the second basis vector b₂ swings, while b₁ stays at (1, 0). As you can see, the recipe c₁b₁ + c₂b₂ that reaches x keeps changing: when b₂ points straight up, x's coordinates are (2, 1.5), and as b₂ tips further left they climb to about (4.68, 2.01). Coordinates are an address relative to a chosen basis, not a property of the point itself.

Worked examples

Example 1

With $\mathcal B = \left\{\binom10, \binom12\right\}$ , find $\vec x$ if $[\vec x]_{\mathcal B} = \binom31$ .

1
The coordinates are the weights, so build $\vec x$ from the basis: $3\binom10 + 1\binom12$ .

$\vec x = \binom42$

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

With $\mathcal B = \left\{\binom10, \binom12\right\}$ and $[\vec x]_{\mathcal B} = \binom31$ , find $\vec x$ .

Problem 2+5 XP

With the same basis, find $[\vec x]_{\mathcal B}$ for $\vec x = \binom42$ .

Problem 3+5 XP

Find the coordinates of $3 + 2t - t^2$ in the basis $\{1, t, t^2\}$ .

Problem 4+5 XP

In the basis $\left\{\binom20, \binom03\right\}$ , find the coordinates of $\binom49$ .

Problem 5+5 XP

Give $P_{\mathcal B}$ for $\mathcal B = \left\{\binom10, \binom12\right\}$ , row by row.

Problem 6+5 XP

A robot vacuum maps a room with its own axes: $\vec b_1 = (2, 1)$ along a wall and $\vec b_2 = (-1, 2)$ perpendicular to it, in meters of room coordinates. Its sensor reports a chair at $[\vec x]_{\mathcal B} = (1.5, 0.5)$ . Where is the chair in room coordinates?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the coordinates of $p(t) = 3 + 2t + 5t^2$ relative to the basis $\mathcal B = \{1 + t,\ t + t^2,\ 1 + t^2\}$ of $\mathbb P_2$ .

Where it's used

GPS, robotics and graphics constantly convert between coordinate frames: the world frame, the camera frame, each joint's frame. Each conversion is a change of coordinates.

Robot Arm KinematicsLin Alg · Robotics Coordinates on a Crystal LatticeLin Alg · Physics

Linearly Independent Sets and Bases The Dimension of a Vector Space