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# Change of basis

In linear algebra, a basis for a vector space of dimension n is a sequence of n vectors α1, ..., αn 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 .

## Preliminary notions

The standard basis for Rn is {e1, ..., en}, where ej = (0, ..., 1, 0, ..., 0) is the element of Rn with 1 in the j-th place and 0s elsewhere.

If T : Rn → Rm is a linear transformation, the m × n matrix of T is the matrix t whose j-th column is T(ej) 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, Rm) of all linear transformations from Rn to Rm is naturally isomorphic to the space Rm × n of m × n matrices over R; that is, a linear transformation T : Rn → Rm 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, ..., αn} be a basis for V, and let {γ1, ..., γn} be any n vectors in W. Then there exists a unique linear transformation T : VW with Tj) = γj for j = 1, ..., n.

This unique T is defined by T(x1α1 + ... + xnαn) = x1γ1 + ... + xnγn. Of course, if {γ1, ..., γn} 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, ..., αn} is a basis for V. By definition, if ξ is a vector in V then ξ = x1α1 + ... + xnαn for a unique choice of scalars x1, ..., xn in R called the coordinates of ξ relative to the ordered basis1, ..., αn}. The vector x = (x1, ..., xn) in Rn is called the coordinate tuple of ξ (relative to this basis). The unique linear map φ : RnV with φ(ej) = αj for j = 1, ..., n is called the coordinate isomorphism for V and the basis {α1, ..., αn}. Thus φ(x) = ξ if and only if ξ = x1α1 + ... + xnαn.

## Change of coordinates

First we examine the question of how the coordinates of a vector ξ in V change when we select another basis. Suppose {α1, ..., αn} and {α'1, ..., α'n} are two ordered bases for V. Let φ1 and φ2 be the corresponding coordinate isomorphisms from Rn to V, i.e. φ1(ej) = αj and φ2(ej) = α'j for j = 1, ..., n. If x = (x1, ..., xn) 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) = φ2-11(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(ej) = φ2-1j), that is, the coordinate n-tuple of αj with respect to the second basis {α'1, ..., α'n}. Thus y = φ2-11(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 : VW is a linear transformation, {α1, ..., αn} is a basis for V and {β1, ..., βm} 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 Rm, and therefore has a matrix t; its j-th column is ψ-1(T(αj)) for j = 1, ..., n. This matrix is called the matrix of T with respect to the ordered bases1, ..., αn} and1, ..., βm}. 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, ..., αn}, 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 : UV and S : VW are linear transformations with matrices s and t, then the matrix of the linear transformation S o T : UW (with respect to the given bases) is st.

### Change of basis

Now we ask what happens to the matrix of T : VW when we change bases in V and W. Let {α1, ..., αn} and {β1, ..., βm} 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 Rm 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 Rm), 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 Rm.

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

(insert standard change-of-basis 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.

## 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, ..., βn} = {α1, ..., αn} 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 × VR which is linear in both arguments. That is, B : V × VR is bilinear if the maps

$v mapsto B\left(v, w\right)$
$v mapsto B\left(w, v\right)$
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

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

If $v = sum_i x_i alpha_i$ and $w = sum_i y_i alpha_i$ are the expressions of vectors v, w with respect to this basis, then the bilinear form is given by

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

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 $alpha\text{'}_1,dots, alpha\text{'}_n$ then the Gram matrix transforms by the matrix congruence

$G\text{'} = P^top G P ,$ .

## Example from mechanics

(this example to be replaced or amended)

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 could formulate this as a matrix, where 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):

$A=begin\left\{pmatrix\right\}$
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:

$R=begin\left\{pmatrix\right\}$
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:

$R^\left\{-1\right\} P$

Applying our matrix A:

$A \left(R^\left\{-1\right\} P\right)$

Changing back to the original basis:

$R \left(A \left(R^\left\{-1\right\} P\right)\right)$

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

$R A R^\left\{-1\right\} P$

And identify the matrix we were looking for:

$R A R^\left\{-1\right\}$

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:

$B=R A R^\left\{T\right\} = begin\left\{pmatrix\right\}$
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:

$B P = begin\left\{pmatrix\right\}$
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 xxyy = 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 real matrices (2 x 2) 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 x 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.