Change of Basis

One of the things I knew well enough to pass an exam on, but not well enough to actually understand what it meant was the concept of changing basis in linear algebra. This is an attempt to remedy that.

A basis for a vector space \(V\) of dimension \(n\) is a sequence of \(n\) vectors \((v_1, v_2, \dots, v_n)\) such that all vectors in \(V\) can be uniquely expressed as linear combinations of the basis vectors. Often calculations in one basis can be simplified if we change our basis to something else.

For example, if we wish to work with a vector \(\vec v\) with respect to the standard basis vectors \(\vec x = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\), \(\vec y = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\) and \(\vec z = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\), we know how to do that. Here's a graph of some random vector \(\vec v\).

vector v wrt the standard basis
A random vector in \(\mathbb R^3\)

What if we want to rotate our coordinate system by some Euler angles \(\alpha\), \(\beta\), and \(\gamma\)? This is what changing the basis really is. To rotate a vector \(\vec v\) given some Euler angles, multiply \(\vec v\) by the following transition matrix \(P\) taken from the \(Z_1 Y_2 X_3\) entry of this table. Source.

\[P = \begin{pmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta \sin \gamma - \sin \alpha \cos \gamma & \cos \alpha \sin \beta \cos \gamma + \sin \alpha \sin \gamma \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta \sin \gamma + \cos \alpha \cos \gamma & \sin \alpha \sin \beta \cos \gamma - \cos \alpha \sin \gamma \\ - \sin \beta & \cos \beta \sin \gamma & \cos \beta \cos \gamma \end{pmatrix}\]

With the Euler angles from the following diagram:

Euler angles
Euler angles

We get a new vector \(v' = P v\) that, with respect to the new coordinate system, is the same as \(\vec v\).

vector v' wrt the new rotated basis
The same vector in a different basis

In the first image, \(\vec x\), \(\vec y\), and \(\vec z\) form the basis for the vector space. In the second image, \(\vec{x'}\), \(\vec{y'}\), and \(\vec{z'}\) form the basis for the new, rotated, vector space.

In general

To convert a vector \(\vec v\) w.r.t. the standard basis to a vector \(\vec{v'}\) w.r.t. a new basis \(\{\vec{u_1}, \vec{u_2}, \dots, \vec{u_n}\}\) we multiply \(\vec v\) by the matrix \(U\) formed by \(U = (\vec{u_1}, \vec{u_2}, \dots, \vec{u_n})\). That is,

\[\vec{v'} = U \cdot \vec v\]

To convert a vector \(\vec v\) w.r.t. a basis \(\{\vec{u_1}, \vec{u_2}, \dots, \vec{u_n}\}\) to the standard basis, we multiply \(\vec v\) by the matrix \(U^{-1}\). That is,

\[\vec{v'} = U^{-1} \cdot \vec v\]

In general, to convert a vector \(\vec v\) w.r.t. a basis \(V\) to a vector \(\vec{v'}\) in a basis \(U\) formed by \(U = (\vec{u_1}, \vec{u_2}, \dots, \vec{u_n})\) and \(V = (\vec{v_1}, \vec{v_2}, \dots, \vec{v_n})\), we multiply \(\vec v\) by the transition matrix \(U^{-1} V\). That is,

\[\vec{v'} = U^{-1} V \cdot \vec v\]

In particular, we can use rotation matrices as transition matrices to rotate our basis by some angle(s). The real trick is to find an appropriate transition matrix, the actual change of basis isn't too bad.