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Chapter 24. Change of Basis

A change of basis converts the coordinate description of a vector from one basis to another. The vector itself does not change. Only the coordinate column changes.

If a vector space has two ordered bases, then the same vector has two coordinate vectors. A change-of-basis matrix gives the linear rule that converts one coordinate vector into the other. In finite-dimensional spaces, this matrix is invertible because both coordinate systems describe the same vector space.

24.1 The Problem

Let VV be a finite-dimensional vector space over a field FF. Let

B=(b1,…,bn) B=(b_1,\ldots,b_n)

and

C=(c1,…,cn) C=(c_1,\ldots,c_n)

be ordered bases of VV.

For a vector v∈Vv\in V, we may know its coordinates in basis BB:

[v]B. [v]_B.

We may want its coordinates in basis CC:

[v]C. [v]_C.

The change-of-basis problem is to compute

[v]C [v]_C

from

[v]B. [v]_B.

The vector vv remains fixed. The coordinate system changes.

24.2 Coordinate Equality

Since BB is a basis, the vector vv has a unique expression

v=x1b1+⋯+xnbn. v=x_1b_1+\cdots+x_nb_n.

Thus

[v]B=[x1⋮xn]. [v]_B= \begin{bmatrix} x_1\\ \vdots\\ x_n \end{bmatrix}.

Since CC is also a basis, the same vector has another expression

v=y1c1+⋯+yncn. v=y_1c_1+\cdots+y_nc_n.

Thus

[v]C=[y1⋮yn]. [v]_C= \begin{bmatrix} y_1\\ \vdots\\ y_n \end{bmatrix}.

The two coordinate columns are usually different, but they describe the same vector.

24.3 Basis Matrices in Rn\mathbb{R}^n

In Rn\mathbb{R}^n, a basis can be stored as a matrix whose columns are the basis vectors.

For

B=(b1,…,bn), B=(b_1,\ldots,b_n),

define

PB=[∣∣∣b1b2⋯bn∣∣∣]. P_B= \begin{bmatrix} |&|&&|\\ b_1&b_2&\cdots&b_n\\ |&|&&| \end{bmatrix}.

Then

v=PB[v]B. v=P_B[v]_B.

Similarly, for

C=(c1,…,cn), C=(c_1,\ldots,c_n),

define

PC=[∣∣∣c1c2⋯cn∣∣∣]. P_C= \begin{bmatrix} |&|&&|\\ c_1&c_2&\cdots&c_n\\ |&|&&| \end{bmatrix}.

Then

v=PC[v]C. v=P_C[v]_C.

Since both expressions represent the same vector,

PB[v]B=PC[v]C. P_B[v]_B=P_C[v]_C.

This equation is the starting point for change of basis.

24.4 Change from BB to CC

From

PB[v]B=PC[v]C, P_B[v]_B=P_C[v]_C,

multiply by PC−1P_C^{-1}:

[v]C=PC−1PB[v]B. [v]_C=P_C^{-1}P_B[v]_B.

The matrix

PC←B=PC−1PB P_{C\leftarrow B}=P_C^{-1}P_B

is called the change-of-basis matrix from BB-coordinates to CC-coordinates.

It satisfies

[v]C=PC←B[v]B. [v]_C=P_{C\leftarrow B}[v]_B.

The arrow notation means “coordinates in BB, converted to coordinates in CC.”

24.5 Change from CC to BB

The reverse change of basis is

[v]B=PB−1PC[v]C. [v]_B=P_B^{-1}P_C[v]_C.

Thus

PB←C=PB−1PC. P_{B\leftarrow C}=P_B^{-1}P_C.

The two change-of-basis matrices are inverses:

PB←C=(PC←B)−1. P_{B\leftarrow C} = (P_{C\leftarrow B})^{-1}.

Changing from BB to CC and then back to BB gives the original coordinate vector.

24.6 Columns of a Change-of-Basis Matrix

The columns of

PC←B P_{C\leftarrow B}

have a direct meaning.

Since

PC←B[v]B=[v]C, P_{C\leftarrow B}[v]_B=[v]_C,

apply this to each basis vector bjb_j.

The BB-coordinate vector of bjb_j is the standard coordinate column eje_j. Hence

PC←Bej=[bj]C. P_{C\leftarrow B}e_j=[b_j]_C.

Therefore the jj-th column of PC←BP_{C\leftarrow B} is

[bj]C. [b_j]_C.

So

PC←B=[∣∣∣[b1]C[b2]C⋯[bn]C∣∣∣]. P_{C\leftarrow B} = \begin{bmatrix} |&|&&|\\ [b_1]_C&[b_2]_C&\cdots&[b_n]_C\\ |&|&&| \end{bmatrix}.

This is often the cleanest definition: the change-of-basis matrix from BB to CC has as its columns the CC-coordinates of the BB-basis vectors.

24.7 Example in R2\mathbb{R}^2

Let

B=([11],[1−1]) B= \left( \begin{bmatrix} 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -1 \end{bmatrix} \right)

and let CC be the standard basis

C=(e1,e2). C=(e_1,e_2).

Then

PB=[111−1],PC=[1001]. P_B= \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}, \qquad P_C= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}.

Therefore

PC←B=PC−1PB=PB=[111−1]. P_{C\leftarrow B} = P_C^{-1}P_B = P_B = \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}.

If

[v]B=[52], [v]_B= \begin{bmatrix} 5\\ 2 \end{bmatrix},

then

[v]C=[111−1][52]=[73]. [v]_C = \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix} \begin{bmatrix} 5\\ 2 \end{bmatrix} = \begin{bmatrix} 7\\ 3 \end{bmatrix}.

Thus the vector with BB-coordinates (5,2)(5,2) has standard coordinates (7,3)(7,3).

24.8 Reverse Example

Using the same basis BB, convert standard coordinates back to BB-coordinates.

We need

PB←C=PB−1PC=PB−1. P_{B\leftarrow C} = P_B^{-1}P_C = P_B^{-1}.

Since

PB=[111−1], P_B= \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix},

we have

PB−1=[1/21/21/2−1/2]. P_B^{-1} = \begin{bmatrix} 1/2&1/2\\ 1/2&-1/2 \end{bmatrix}.

For

[v]C=[73], [v]_C= \begin{bmatrix} 7\\ 3 \end{bmatrix},

we compute

[v]B=[1/21/21/2−1/2][73]=[52]. [v]_B = \begin{bmatrix} 1/2&1/2\\ 1/2&-1/2 \end{bmatrix} \begin{bmatrix} 7\\ 3 \end{bmatrix} = \begin{bmatrix} 5\\ 2 \end{bmatrix}.

The two computations are inverse processes.

24.9 Change Between Two Nonstandard Bases

Let

B=([11],[1−1]),C=([21],[11]). B= \left( \begin{bmatrix} 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -1 \end{bmatrix} \right), \qquad C= \left( \begin{bmatrix} 2\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 1 \end{bmatrix} \right).

Then

PB=[111−1],PC=[2111]. P_B= \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix}, \qquad P_C= \begin{bmatrix} 2&1\\ 1&1 \end{bmatrix}.

The change-of-basis matrix from BB to CC is

PC←B=PC−1PB. P_{C\leftarrow B} = P_C^{-1}P_B.

Since

PC−1=[1−1−12], P_C^{-1} = \begin{bmatrix} 1&-1\\ -1&2 \end{bmatrix},

we get

PC←B=[1−1−12][111−1]=[021−3]. P_{C\leftarrow B} = \begin{bmatrix} 1&-1\\ -1&2 \end{bmatrix} \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix} = \begin{bmatrix} 0&2\\ 1&-3 \end{bmatrix}.

Therefore

[v]C=[021−3][v]B. [v]_C= \begin{bmatrix} 0&2\\ 1&-3 \end{bmatrix} [v]_B.

If

[v]B=[41], [v]_B= \begin{bmatrix} 4\\ 1 \end{bmatrix},

then

[v]C=[021−3][41]=[21]. [v]_C= \begin{bmatrix} 0&2\\ 1&-3 \end{bmatrix} \begin{bmatrix} 4\\ 1 \end{bmatrix} = \begin{bmatrix} 2\\ 1 \end{bmatrix}.

Check the vector itself:

v=4[11]+1[1−1]=[53]. v=4 \begin{bmatrix} 1\\ 1 \end{bmatrix} + 1 \begin{bmatrix} 1\\ -1 \end{bmatrix} = \begin{bmatrix} 5\\ 3 \end{bmatrix}.

Also,

2[21]+1[11]=[53]. 2 \begin{bmatrix} 2\\ 1 \end{bmatrix} + 1 \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} 5\\ 3 \end{bmatrix}.

Both coordinate columns describe the same vector.

24.10 The Identity Transformation

A change of basis is not necessarily a transformation of the vector space. Often it is the identity transformation written in two coordinate systems.

The vector vv remains the same. The coordinate column changes from [v]B[v]_B to [v]C[v]_C.

This distinction matters because the same matrix notation can represent two different ideas:

Matrix useMeaning
Transformation matrixSends one vector to another vector
Change-of-basis matrixSends one coordinate description to another coordinate description

In change of basis, the object is fixed and the representation changes.

24.11 Abstract Vector Spaces

The formula using PBP_B and PCP_C assumes that vectors are already written in standard coordinates, as in Rn\mathbb{R}^n. For abstract vector spaces, the column interpretation still works.

Let

B=(b1,…,bn) B=(b_1,\ldots,b_n)

and

C=(c1,…,cn) C=(c_1,\ldots,c_n)

be bases of VV.

To build PC←BP_{C\leftarrow B}, write each bjb_j in the CC-basis:

bj=a1jc1+⋯+anjcn. b_j=a_{1j}c_1+\cdots+a_{nj}c_n.

Then the jj-th column of PC←BP_{C\leftarrow B} is

[bj]C=[a1j⋮anj]. [b_j]_C= \begin{bmatrix} a_{1j}\\ \vdots\\ a_{nj} \end{bmatrix}.

Thus

PC←B=(aij). P_{C\leftarrow B} = (a_{ij}).

This method works for polynomials, matrices, functions, and any finite-dimensional vector space.

24.12 Polynomial Example

Let P2P_2 be the space of polynomials of degree at most 22.

Let

B=(1,x,x2) B=(1,x,x^2)

and

C=(1,1+x,1+x+x2). C=(1,1+x,1+x+x^2).

We construct PC←BP_{C\leftarrow B}. Its columns are

[1]C,[x]C,[x2]C. [1]_C,\qquad [x]_C,\qquad [x^2]_C.

First,

1=1⋅1+0(1+x)+0(1+x+x2), 1=1\cdot 1+0(1+x)+0(1+x+x^2),

so

[1]C=[100]. [1]_C= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}.

Next,

x=−1⋅1+1(1+x)+0(1+x+x2), x=-1\cdot 1+1(1+x)+0(1+x+x^2),

so

[x]C=[−110]. [x]_C= \begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix}.

Finally,

x2=0⋅1−1(1+x)+1(1+x+x2), x^2=0\cdot 1-1(1+x)+1(1+x+x^2),

so

[x2]C=[0−11]. [x^2]_C= \begin{bmatrix} 0\\ -1\\ 1 \end{bmatrix}.

Therefore

PC←B=[1−1001−1001]. P_{C\leftarrow B} = \begin{bmatrix} 1&-1&0\\ 0&1&-1\\ 0&0&1 \end{bmatrix}.

For

p(x)=3−5x+2x2, p(x)=3-5x+2x^2,

we have

[p]B=[3−52]. [p]_B= \begin{bmatrix} 3\\ -5\\ 2 \end{bmatrix}.

Thus

[p]C=[1−1001−1001][3−52]=[8−72]. [p]_C = \begin{bmatrix} 1&-1&0\\ 0&1&-1\\ 0&0&1 \end{bmatrix} \begin{bmatrix} 3\\ -5\\ 2 \end{bmatrix} = \begin{bmatrix} 8\\ -7\\ 2 \end{bmatrix}.

So

p(x)=8−7(1+x)+2(1+x+x2). p(x)=8-7(1+x)+2(1+x+x^2).

24.13 Matrix Example

Let M2×2(R)M_{2\times2}(\mathbb{R}) have the standard ordered basis

B=(E11,E12,E21,E22). B=(E_{11},E_{12},E_{21},E_{22}).

Let

C=(A1,A2,A3,A4), C=(A_1,A_2,A_3,A_4),

where

A1=E11,A2=E11+E12,A3=E21,A4=E21+E22. A_1=E_{11}, \quad A_2=E_{11}+E_{12}, \quad A_3=E_{21}, \quad A_4=E_{21}+E_{22}.

To compute coordinates in CC, write a matrix

M=[abcd] M= \begin{bmatrix} a&b\\ c&d \end{bmatrix}

as

M=αA1+βA2+γA3+δA4. M = \alpha A_1+\beta A_2+\gamma A_3+\delta A_4.

Then

αA1+βA2+γA3+δA4=[α+ββγ+δδ]. \alpha A_1+\beta A_2+\gamma A_3+\delta A_4 = \begin{bmatrix} \alpha+\beta&\beta\\ \gamma+\delta&\delta \end{bmatrix}.

Matching entries gives

β=b,α=a−b,δ=d,γ=c−d. \beta=b, \qquad \alpha=a-b, \qquad \delta=d, \qquad \gamma=c-d.

Therefore

[M]C=[a−bbc−dd]. [M]_C= \begin{bmatrix} a-b\\ b\\ c-d\\ d \end{bmatrix}.

This illustrates the same rule in a vector space whose vectors are matrices.

24.14 Change of Basis and Linear Maps

Let

T:V→V T:V\to V

be a linear transformation. Suppose its matrix in basis BB is

[T]B. [T]_B.

If CC is another basis, the matrix of the same transformation in basis CC is related by a change of basis.

Let

S=PB←C. S=P_{B\leftarrow C}.

This matrix converts CC-coordinates into BB-coordinates:

[v]B=S[v]C. [v]_B=S[v]_C.

Apply the transformation in BB-coordinates:

[T(v)]B=[T]B[v]B. [T(v)]_B=[T]_B[v]_B.

Convert back to CC-coordinates:

[T(v)]C=S−1[T(v)]B. [T(v)]_C=S^{-1}[T(v)]_B.

Substitute:

[T(v)]C=S−1[T]BS[v]C. [T(v)]_C = S^{-1}[T]_B S [v]_C.

Therefore

[T]C=S−1[T]BS. [T]_C=S^{-1}[T]_B S.

This is the similarity formula.

24.15 Similar Matrices

Two square matrices AA and BB are similar if there exists an invertible matrix SS such that

B=S−1AS. B=S^{-1}AS.

Similar matrices represent the same linear transformation in different bases.

They may look different, but they share important structural properties, including determinant, trace, rank, characteristic polynomial, eigenvalues, and invertibility.

Similarity is the matrix form of changing the coordinate system for a linear operator.

24.16 Diagonalization as Change of Basis

Diagonalization is a special change of basis.

A matrix AA is diagonalizable if there is a basis made of eigenvectors of the corresponding linear transformation. In that basis, the matrix becomes diagonal.

If

S=[∣∣∣v1v2⋯vn∣∣∣] S= \begin{bmatrix} |&|&&|\\ v_1&v_2&\cdots&v_n\\ |&|&&| \end{bmatrix}

has eigenvectors as columns, and

Avi=λivi, Av_i=\lambda_i v_i,

then

S−1AS=[λ10⋯00λ2⋯0⋮⋮⋱⋮00⋯λn]. S^{-1}AS= \begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}.

This is not a different transformation. It is the same transformation described in an eigenvector basis.

24.17 Common Sources of Confusion

The phrase “change-of-basis matrix” is used in more than one convention. Some texts define the matrix from new coordinates to old coordinates. Others define the matrix from old coordinates to new coordinates.

The safest method is to write the equation explicitly.

Desired conversionMatrix
BB-coordinates to standard coordinatesPBP_B
Standard coordinates to BB-coordinatesPB−1P_B^{-1}
BB-coordinates to CC-coordinatesPC−1PBP_C^{-1}P_B
CC-coordinates to BB-coordinatesPB−1PCP_B^{-1}P_C

Always check the direction by applying the matrix to a coordinate column.

24.18 Change of Basis as Relabeling

Geometrically, changing basis relabels the same vector with respect to different axes.

In R2\mathbb{R}^2, standard coordinates measure horizontal and vertical displacement. A nonstandard basis may use slanted axes. The point or vector in the plane remains fixed, but the numbers used to reach it along the chosen basis directions change.

This explains why change of basis is invertible. A valid basis gives a complete coordinate system. No information is lost by changing from one basis to another.

24.19 Computational Procedure

To change coordinates from basis BB to basis CC in Rn\mathbb{R}^n:

StepOperation
1Form PBP_B from the columns b1,…,bnb_1,\ldots,b_n
2Form PCP_C from the columns c1,…,cnc_1,\ldots,c_n
3Compute PC←B=PC−1PBP_{C\leftarrow B}=P_C^{-1}P_B
4Multiply [v]C=PC←B[v]B[v]_C=P_{C\leftarrow B}[v]_B

For an abstract vector space, replace the basis matrices by coordinate columns. Express each BB-basis vector in the CC-basis, and place those coordinate columns into the matrix.

24.20 Summary

Change of basis converts coordinates from one ordered basis to another. It does not change the vector itself.

The key ideas are:

ConceptMeaning
Basis matrixMatrix whose columns are basis vectors
PB[v]BP_B[v]_BConvert BB-coordinates to standard coordinates
PC−1PBP_C^{-1}P_BConvert BB-coordinates to CC-coordinates
PC←BP_{C\leftarrow B}Change-of-basis matrix from BB to CC
Columns of PC←BP_{C\leftarrow B}CC-coordinates of the BB-basis vectors
Similar matricesSame linear operator in different bases
DiagonalizationChange to an eigenvector basis

A basis gives coordinates. A change of basis changes coordinates. The underlying vector or linear transformation remains the same, while its numerical representation changes.