Change of Basis

Main idea

Changing basis rewrites the same vector's coordinates in a new basis, and one matrix does all the work: $P_{\mathcal C\leftarrow\mathcal B}$ , whose columns are the $\mathcal B$ vectors written in $\mathcal C$ -coordinates. Multiply by it to go from $\mathcal B$ to $\mathcal C$ , and by its inverse to come back.

[\vec x]_{\mathcal C} = {\htmlData{sym=0}{P_{\mathcal C\leftarrow\mathcal B}}}\,[\vec x]_{\mathcal B}, \qquad {\htmlData{sym=0}{P_{\mathcal C\leftarrow\mathcal B}}} = \begin{bmatrix}[\vec b_1]_{\mathcal C} & \cdots & [\vec b_n]_{\mathcal C}\end{bmatrix}

What the symbols mean

$P_{\mathcal C\leftarrow\mathcal B}$
Change-of-basis matrix
Translates B-coordinates into C-coordinates. Its inverse goes the other way.

Explanation

A change of basis is simply translating a vector's coordinates from one basis into another, like converting an address from one map's grid to another's. The vector itself never moves; only its description changes. That's useful because some problems are easy in one basis and messy in another, and you want to be able to switch freely. All we're really saying in $[\vec x]_{\mathcal C} = P_{\mathcal C\leftarrow\mathcal B}[\vec x]_{\mathcal B}$ is: start with $\mathcal B$ -coordinates (on the right), multiply by one matrix, and get $\mathcal C$ -coordinates (on the left). The arrow in $P_{\mathcal C\leftarrow\mathcal B}$ reads right to left (from $\mathcal B$ , into $\mathcal C$ ), and its columns are just the $\mathcal B$ vectors written in $\mathcal C$ -coordinates. It's a lot of subscripts for one simple idea; bear with it.

Why those columns? Think about it this way. If $\vec x = c_1\vec b_1 + \cdots + c_n\vec b_n$ , then $[\vec x]_{\mathcal C} = c_1[\vec b_1]_{\mathcal C} + \cdots + c_n[\vec b_n]_{\mathcal C}$ , because coordinates respect sums and multiples. That's a combination of the columns $[\vec b_j]_{\mathcal C}$ with weights $c_1, \dots, c_n$ , which is exactly a matrix times $[\vec x]_{\mathcal B}$ .

Into the standard basis it's just $P_{\mathcal B}$ (the basis vectors as columns), the same matrix from the coordinate systems lesson.

Between two nonstandard bases, go through the standard one: $P_{\mathcal C\leftarrow\mathcal B} = P_{\mathcal C}^{-1}P_{\mathcal B}$ .

A common mistake. It's tempting to write that product the other way around, $P_{\mathcal B}^{-1}P_{\mathcal C}$ , because $\mathcal B$ comes first in the sentence "from $\mathcal B$ to $\mathcal C$ ." It isn't: read the product right to left, the way it acts on a vector. $P_{\mathcal B}$ goes first, turning $\mathcal B$ -coordinates into standard coordinates, and then $P_{\mathcal C}^{-1}$ turns those into $\mathcal C$ -coordinates.

Matrices in a new basis. A transformation with matrix $A$ in standard coordinates has matrix $P^{-1}AP$ in the basis given by $P$ 's columns. Read it right to left again: $P$ turns new coordinates into standard ones, $A$ does the transformation, and $P^{-1}$ turns the answer back into new coordinates. Diagonalization (next unit) is choosing the basis that makes this diagonal.

There's an animation further down that shows two different recipes landing on the same point.

See it

Two recipes for the same point x as it circles: one steps along B = {(1, 1), (1, −1)}, the other along C = {(1, 0), (1, 2)}, and both always land on x. When x = (2.5, 0), for instance, its B-coordinates are (1.25, 1.25) and its C-coordinates are (2.5, 0). The change-of-coordinates matrix from B to C, [[1/2, 3/2], [1/2, −1/2]], turns one recipe straight into the other: multiply it by (1.25, 1.25) and out comes (2.5, 0).

Worked examples

Example 1

With $\mathcal B = \left\{\binom11, \binom{1}{-1}\right\}$ and $[\vec x]_{\mathcal B} = \binom21$ , find $\vec x$ in standard coordinates.

1
Going into the standard basis just takes $P_{\mathcal B}$ , whose columns are the basis vectors: $P_{\mathcal B}[\vec x]_{\mathcal B} = 2\binom11 + 1\binom{1}{-1}$ .

$\binom31$

Practice

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

Problem 1+5 XP

With $\mathcal B = \left\{\binom11, \binom{1}{-1}\right\}$ and $[\vec x]_{\mathcal B} = \binom21$ , find $\vec x$ .

Problem 2+5 XP

With the same basis, find $[\vec x]_{\mathcal B}$ for $\vec x = \binom31$ .

Problem 3+5 XP

How is $P_{\mathcal B\leftarrow\mathcal C}$ related to $P_{\mathcal C\leftarrow\mathcal B}$ ?

Problem 4+5 XP

$\mathcal B = \left\{\binom20, \binom01\right\}$ and $\mathcal C = \left\{\binom10, \binom02\right\}$ . Give $P_{\mathcal C\leftarrow\mathcal B}$ , row by row.

Problem 5+5 XP

Using that matrix, convert $[\vec x]_{\mathcal B} = \binom14$ to $\mathcal C$ -coordinates.

Problem 6+5 XP

Two delivery robots share a map. Robot 1 records positions in the basis $\mathcal B = \left\{\binom11, \binom{1}{-1}\right\}$ and robot 2 in $\mathcal C = \left\{\binom10, \binom12\right\}$ . Robot 1 reports an obstacle at $[\vec x]_{\mathcal B} = \binom21$ . What are its coordinates for robot 2?

Challenge problem+25 XP

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

In $\mathbb P_2$ , let $\mathcal B = \{1, t, t^2\}$ and $\mathcal C = \{1,\ t - 1,\ (t - 1)^2\}$ . Find $P_{\mathcal C\leftarrow\mathcal B}$ , row by row.

Where it's used

Animation rigs and robots chain changes of basis from one joint frame to the next; color spaces (RGB to YUV) are changes of basis too.

Robot Arm KinematicsLin Alg · Robotics 3D TransformationsLin Alg · Game Development

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