Change of basis

In linear algebra, a basis for a vector space of dimension n is a sequence of n vectors α₁, ..., α_n with the property that every vector in the space can be expressed uniquely as a linear combination of the basis vectors. Since it is often desirable to work with more than one basis for a vector space, it is of fundamental importance in linear algebra to be able to easily transform coordinate-wise representations of vectors and linear transformations taken with respect to one basis to their equivalent representations with respect to another basis. Such a transformation is called a change of basis.

Although the terminology of vector spaces is used below and the symbol R can be taken to mean the field of real numbers, the results discussed hold whenever R is a commutative ring and vector space is everywhere replaced with free R-module.

Preliminary notions

The standard basis for Rⁿ is {e₁, ..., e_n}, where e_j is the element of Rⁿ with 1 in the j-th place and 0s elsewhere.

If T : Rⁿ → R^m is a linear transformation, the m × n matrix of T is the matrix t whose j-th column is T(e_j) for j = 1, ..., n. In this case we have T(x) = tx for all x in Rⁿ, where we regard x as a column vector and the multiplication on the right side is matrix multiplication. It is a basic fact in linear algebra that the vector space Hom(Rⁿ, R^m) of all linear transformations from Rⁿ to R^m is naturally isomorphic to the space R^{m × n} of m × n matrices over R; that is, a linear transformation T : Rⁿ → R^m is for all intents and purposes equivalent to its matrix t.

We will also make use of the following simple observation.

Theorem Let V and W be vector spaces, let {α₁, ..., α_n} be a basis for V, and let {γ₁, ..., γ_n} be any n vectors in W. Then there exists a unique linear transformation T : V → W with T(α_j) = γ_j for j = 1, ..., n.

This unique T is defined by T(x₁α₁ + ... + x_nα_n) = x₁γ₁ + ... + x_nγ_n. Of course, if {γ₁, ..., γ_n} happens to be a basis for W, then T is bijective as well as linear; in other words, T is an isomorphism. If in this case we also have W = V, then T is said to be an automorphism.

Now let V be a vector space over R and suppose {α₁, ..., α_n} is a basis for V. By definition, if ξ is a vector in V then ξ = x₁α₁ + ... + x_nα_n for a unique choice of scalars x₁, ..., x_n in R called the coordinates of ξ relative to the ordered basis {α₁, ..., α_n}. The vector x = (x₁, ..., x_n) in Rⁿ is called the coordinate tuple of ξ (relative to this basis). The unique linear map φ : Rⁿ → V with φ(e_j) = α_j for j = 1, ..., n is called the coordinate isomorphism for V and the basis {α₁, ..., α_n}. Thus φ(x) = ξ if and only if ξ = x₁α₁ + ... + x_nα_n.

Matrix of a set of vectors

A set of vectors can be represented by a matrix whose columns are the components of each vector of the set. As a basis is a set of vectors, a basis can be given by a matrix of this kind. Later it will be shown that the change of basis of any object of the space is related to this matrix. For example vectors change with its inverse (and they are therefore called contravariant objects).

Change of coordinates of a vector

First we examine the question of how the coordinates of a vector ξ in V change when we select another basis.

Two dimensions

This means that given a matrix M whose columns are the vectors of the new basis of the space (new basis matrix), the new coordinates for a column vector v are given by the matrix product M^-1.v. For this reason, it is said that normal vectors are contravariant objects.

Any group of vectors can be represented by a matrix in which its columns are the coordinates of every vector. As an example in dimension 2, a pair of vectors rotated 45 degrees from the basis of the space rotation looks like this:

$R=\begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$

If we want to change any vector of the space to this new basis, we only need to left-multiply its components by the inverse of this matrix.

Three dimensions

For example, be a new basis given by its Euler angles. The matrix of the basis will have as columns the components of each vector. Therefore, this matrix will be (See Euler angles article):

$[\mathbf{R}] = \begin{bmatrix} \mathrm{c}_\alpha \, \mathrm{c}_\gamma - \mathrm{s}_\alpha \, \mathrm{c}_\beta \, \mathrm{s}_\gamma & -\mathrm{c}_\alpha \, \mathrm{s}_\gamma - \mathrm{s}_\alpha \, \mathrm{c}_\beta \, \mathrm{c}_\gamma & \mathrm{s}_\beta \, \mathrm{s}_\alpha \\ \mathrm{s}_\alpha \, \mathrm{c}_\gamma + \mathrm{c}_\alpha \, \mathrm{c}_\beta \, \mathrm{s}_\gamma & -\mathrm{s}_\alpha \, \mathrm{s}_\gamma + \mathrm{c}_\alpha \, \mathrm{c}_\beta \, \mathrm{c}_\gamma & -\mathrm{s}_\beta \, \mathrm{c}_\alpha \\ \mathrm{s}_\beta \, \mathrm{s}_\gamma & \mathrm{s}_\beta \, \mathrm{c}_\gamma & \mathrm{c}_\beta \end{bmatrix} .$

Again, any vector of the space can be changed to this new basis by left-multiplying its components by the inverse of this matrix.

General case

Suppose {α₁, ..., α_n} and {α'₁, ..., α'_n} are two ordered bases for V. Let φ₁ and φ₂ be the corresponding coordinate isomorphisms (linear maps) from Rⁿ to V, i.e. φ₁(e_j) = α_j and φ₂(e_j) = α'_j for j = 1, ..., n.

If x = (x₁, ..., x_n) is the coordinate n-tuple of ξ with respect to the first basis, so that ξ = φ₁(x), then the coordinate tuple of ξ with respect to the second basis is φ₂^-1(ξ) = φ₂^-1(φ₁(x)). Now the map φ₂^-1 o φ₁ is an automorphism on Rⁿ and therefore has a matrix p. Moreover, the j-th column of p is φ₂^-1 o φ₁(e_j) = φ₂^-1(α_j), that is, the coordinate n-tuple of α_j with respect to the second basis {α'₁, ..., α'_n}. Thus y = φ₂^-1(φ₁(x)) = px is the coordinate n-tuple of ξ with respect to the basis {α'₁, ..., α'_n}.

The matrix of a linear transformation

Now suppose T : V → W is a linear transformation, {α₁, ..., α_n} is a basis for V and {β₁, ..., β_m} is a basis for W. Let φ and ψ be the coordinate isomorphisms for V and W, respectively, relative to the given bases. Then the map T₁ = ψ^-1 o T o φ is a linear transformation from Rⁿ to R^m, and therefore has a matrix t; its j-th column is ψ^-1(T(α_j)) for j = 1, ..., n. This matrix is called the matrix of T with respect to the ordered bases {α₁, ..., α_n} and {β₁, ..., β_m}. If η = T(ξ) and y and x are the coordinate tuples of η and ξ, then y = ψ^-1(T(φ(x))) = tx. Conversely, if ξ is in V and x = φ^-1(ξ) is the coordinate tuple of ξ with respect to {α₁, ..., α_n}, and we set y = tx and η = ψ(y), then η = ψ(T₁(x)) = T(ξ). That is, if ξ is in V and η is in W and x and y are their coordinate tuples, then y = tx if and only if η = T(ξ).

Theorem Suppose U, V and W are vector spaces of finite dimension and an ordered basis is chosen for each. If T : U → V and S : V → W are linear transformations with matrices s and t, then the matrix of the linear transformation S o T : U → W (with respect to the given bases) is st.

Change of basis

Now we ask what happens to the matrix of T : V → W when we change bases in V and W. Let {α₁, ..., α_n} and {β₁, ..., β_m} be ordered bases for V and W respectively, and suppose we are given a second pair of bases {α'₁, ..., α'_n} and {β'₁, ..., β'_m}. Let φ₁ and φ₂ be the coordinate isomorphisms taking the usual basis in Rⁿ to the first and second bases for V, and let ψ₁ and ψ₂ be the isomorphisms taking the usual basis in R^m to the first and second bases for W.

Let T₁ = ψ₁^-1 o T o φ₁, and T₂ = ψ₂^-1 o T o φ₂ (both maps taking Rⁿ to R^m), and let t₁ and t₂ be their respective matrices. Let p and q be the matrices of the change-of-coordinates automorphisms φ₂^-1 o φ₁ on Rⁿ and ψ₂^-1 o ψ₁ on R^m.

The relationships of these various maps to one another are illustrated in the following commutative diagram.

Since we have T₂ = ψ₂^-1 o T o φ₂ = (ψ₂^-1 o ψ₁) o T₁ o (φ₁^-1 o φ₂), and since composition of linear maps corresponds to matrix multiplication, it follows that

t₂ = q t₁ p^-1.

Given that the change of basis has once the basis matrix and once its inverse, this objects are said to be 1-co, 1-contra-variants

The matrix of an endomorphism

An important case of the matrix of a linear transformation is that of an endomorphism, that is, a linear map from a vector space V to itself: that is, the case that W = V. We can naturally take {β₁, ..., β_n} = {α₁, ..., α_n} and {β'₁, ..., β'_m} = {α'₁, ..., α'_n}. The matrix of the linear map T is necessarily square.

Change of basis

We apply the same change of basis, so that q = p and the change of basis formula becomes

t₂ = p t₁ p^-1.

In this situation the invertible matrix p is called a change-of-basis matrix for the vector space V, and the equation above says that the matrices t₁ and t₂ are similar.

The matrix of a bilinear form

A bilinear form on a vector space V over a field R is a mapping V × V → R which is linear in both arguments. That is, B : V × V → R is bilinear if the maps

$v \mapsto B(v, w)$

$v \mapsto B(w, v)$

are linear for each w in V. This definition applies equally well to modules over a commutative ring with linear maps being module homomorphisms.

The Gram matrix G attached to a basis $\alpha_1,\dots, \alpha_n$ is defined by

$G_{i,j} = B(\alpha_i,\alpha_j). \,$

If

v =	∑	xiαi
	i

and

w =	∑	yiαi
	i

are the expressions of vectors v, w with respect to this basis, then the bilinear form is given by

$B(v,w) = v^\top G w. \,$

The matrix will be symmetric if the bilinear form B is a symmetric bilinear form.

Change of basis

If P is the invertible matrix representing a change of basis from $\alpha_1,\dots, \alpha_n$ to $\alpha'_1,\dots, \alpha'_n$ then the Gram matrix transforms by the matrix congruence

$G' = P^\top G P. \,$

Example from mechanics

Let's say we have a train rolling on rails. Using a cartesian coordinate system, let the rail be headed straight in the X-direction (east, on most maps). Now, if we push the train in the X-direction, it will move, but if we try to push the train in the Y-direction (north), it won't be able to move (without derailing).

We can work with both vectors at the same time using a matrix where the components of both are its columns. Here, the first column shows acceleration of the train when pushed in the X-direction =(1,0), and the second column shows the acceleration of the train when pushed in the Y-direction =(0,0):

$A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$

Now, let's say that we want the rail to be headed in northeasterly direction (45 degrees on the compass).

How should our matrix look now? If we push the train in the X-direction, it should move a little in the X-direction, but it should also move in the Y-direction.

A basis change lets us find the matrix B which describes the movement of the train on a northeasterly rail.

All we need to do is to change basis to a basis where the X-axis is in the direction of the rails, multiply with our matrix A, and then change back to the original basis.

A rotation matrix for a 45 degree rotation looks like this:

$R=\begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$

Let the direction we're pushing the train in be P. Putting P in our new basis:

R ^{− 1}P

Applying our matrix A:

A(R ^{− 1}P)

Changing back to the original basis:

R(A(R ^{− 1}P))

And using the matrix multiplication laws, we can remove the parenthesis:

RAR ^{− 1}P

And identify the matrix we were looking for:

RAR ^{− 1}

And by remembering that the inverse of a rotation matrix is simpy its transpose (this step isn't really necessary, but transpose is a quicker operation to do by hand than finding the inverse), our final answer is:

$B=R A R^{T} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}$

We can now see (by looking at the first column of the matrix) that if we push the train in the X-direction, it will move in the direction of the rail .

If we try to push in the north-westerly direction, the train will not move:

$B P = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix} \times \begin{pmatrix} -1/2\\ 1/2 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}$

So the matrix we found seems to do the trick.

Important instances

In abstract vector space theory the change of basis concept is innocuous; it seems to add little to science. Yet there are cases in associative algebras where a change of basis is sufficient to turn a caterpillar into a butterfly, figuratively speaking:

• In the split-complex number plane there is an alternative “diagonal basis”. The standard hyperbola xx − yy = 1 becomes xy = 1 after the change of basis. Transformations of the plane that leave the hyperbolae in place correspond to each other, modulo a change of basis. The contextual difference is profound enough to then separate Lorentz boost from squeeze mapping. A panoramic view of the literature of these mappings can be taken using the underlying change of basis.

• With the 2 × 2 real matrices one finds the beginning of a catalogue of linear algebras due to Arthur Cayley. His associate James Cockle put forward in 1849 his algebra of coquaternions or split-quaternions, which are the same algebra as the 2 × 2 real matrices, just laid out on a different matrix basis. Once again it is the concept of change of basis that synthesizes Cayley’s matrix algebra and Cockle’s coquaternions.

• A change of basis turns a 2 × 2 complex matrix into a biquaternion.

See also

• integral transform, the continuous analogue of change of basis.

External links

• MIT Linear Algebra Lecture on Change of Bases at Videolectures.net, from MIT OpenCourseWare