# Chapter 24. Change of Basis

# 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 \(V\) be a finite-dimensional vector space over a field \(F\). Let

$$
B=(b_1,\ldots,b_n)
$$

and

$$
C=(c_1,\ldots,c_n)
$$

be ordered bases of \(V\).

For a vector \(v\in V\), we may know its coordinates in basis \(B\):

$$
[v]_B.
$$

We may want its coordinates in basis \(C\):

$$
[v]_C.
$$

The change-of-basis problem is to compute

$$
[v]_C
$$

from

$$
[v]_B.
$$

The vector \(v\) remains fixed. The coordinate system changes.

## 24.2 Coordinate Equality

Since \(B\) is a basis, the vector \(v\) has a unique expression

$$
v=x_1b_1+\cdots+x_nb_n.
$$

Thus

$$
[v]_B=
\begin{bmatrix}
x_1\\
\vdots\\
x_n
\end{bmatrix}.
$$

Since \(C\) is also a basis, the same vector has another expression

$$
v=y_1c_1+\cdots+y_nc_n.
$$

Thus

$$
[v]_C=
\begin{bmatrix}
y_1\\
\vdots\\
y_n
\end{bmatrix}.
$$

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

## 24.3 Basis Matrices in \(\mathbb{R}^n\)

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

For

$$
B=(b_1,\ldots,b_n),
$$

define

$$
P_B=
\begin{bmatrix}
|&|&&|\\
b_1&b_2&\cdots&b_n\\
|&|&&|
\end{bmatrix}.
$$

Then

$$
v=P_B[v]_B.
$$

Similarly, for

$$
C=(c_1,\ldots,c_n),
$$

define

$$
P_C=
\begin{bmatrix}
|&|&&|\\
c_1&c_2&\cdots&c_n\\
|&|&&|
\end{bmatrix}.
$$

Then

$$
v=P_C[v]_C.
$$

Since both expressions represent the same vector,

$$
P_B[v]_B=P_C[v]_C.
$$

This equation is the starting point for change of basis.

## 24.4 Change from \(B\) to \(C\)

From

$$
P_B[v]_B=P_C[v]_C,
$$

multiply by \(P_C^{-1}\):

$$
[v]_C=P_C^{-1}P_B[v]_B.
$$

The matrix

$$
P_{C\leftarrow B}=P_C^{-1}P_B
$$

is called the change-of-basis matrix from \(B\)-coordinates to \(C\)-coordinates.

It satisfies

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

The arrow notation means “coordinates in \(B\), converted to coordinates in \(C\).”

## 24.5 Change from \(C\) to \(B\)

The reverse change of basis is

$$
[v]_B=P_B^{-1}P_C[v]_C.
$$

Thus

$$
P_{B\leftarrow C}=P_B^{-1}P_C.
$$

The two change-of-basis matrices are inverses:

$$
P_{B\leftarrow C} =
(P_{C\leftarrow B})^{-1}.
$$

Changing from \(B\) to \(C\) and then back to \(B\) gives the original coordinate vector.

## 24.6 Columns of a Change-of-Basis Matrix

The columns of

$$
P_{C\leftarrow B}
$$

have a direct meaning.

Since

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

apply this to each basis vector \(b_j\).

The \(B\)-coordinate vector of \(b_j\) is the standard coordinate column \(e_j\). Hence

$$
P_{C\leftarrow B}e_j=[b_j]_C.
$$

Therefore the \(j\)-th column of \(P_{C\leftarrow B}\) is

$$
[b_j]_C.
$$

So

$$
P_{C\leftarrow B} =
\begin{bmatrix}
|&|&&|\\
[b_1]_C&[b_2]_C&\cdots&[b_n]_C\\
|&|&&|
\end{bmatrix}.
$$

This is often the cleanest definition: the change-of-basis matrix from \(B\) to \(C\) has as its columns the \(C\)-coordinates of the \(B\)-basis vectors.

## 24.7 Example in \(\mathbb{R}^2\)

Let

$$
B=
\left(
\begin{bmatrix}
1\\
1
\end{bmatrix},
\begin{bmatrix}
1\\
-1
\end{bmatrix}
\right)
$$

and let \(C\) be the standard basis

$$
C=(e_1,e_2).
$$

Then

$$
P_B=
\begin{bmatrix}
1&1\\
1&-1
\end{bmatrix},
\qquad
P_C=
\begin{bmatrix}
1&0\\
0&1
\end{bmatrix}.
$$

Therefore

$$
P_{C\leftarrow B} =
P_C^{-1}P_B =
P_B =
\begin{bmatrix}
1&1\\
1&-1
\end{bmatrix}.
$$

If

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

then

$$
[v]_C =
\begin{bmatrix}
1&1\\
1&-1
\end{bmatrix}
\begin{bmatrix}
5\\
2
\end{bmatrix} =
\begin{bmatrix}
7\\
3
\end{bmatrix}.
$$

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

## 24.8 Reverse Example

Using the same basis \(B\), convert standard coordinates back to \(B\)-coordinates.

We need

$$
P_{B\leftarrow C} =
P_B^{-1}P_C =
P_B^{-1}.
$$

Since

$$
P_B=
\begin{bmatrix}
1&1\\
1&-1
\end{bmatrix},
$$

we have

$$
P_B^{-1} =
\begin{bmatrix}
1/2&1/2\\
1/2&-1/2
\end{bmatrix}.
$$

For

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

we compute

$$
[v]_B =
\begin{bmatrix}
1/2&1/2\\
1/2&-1/2
\end{bmatrix}
\begin{bmatrix}
7\\
3
\end{bmatrix} =
\begin{bmatrix}
5\\
2
\end{bmatrix}.
$$

The two computations are inverse processes.

## 24.9 Change Between Two Nonstandard Bases

Let

$$
B=
\left(
\begin{bmatrix}
1\\
1
\end{bmatrix},
\begin{bmatrix}
1\\
-1
\end{bmatrix}
\right),
\qquad
C=
\left(
\begin{bmatrix}
2\\
1
\end{bmatrix},
\begin{bmatrix}
1\\
1
\end{bmatrix}
\right).
$$

Then

$$
P_B=
\begin{bmatrix}
1&1\\
1&-1
\end{bmatrix},
\qquad
P_C=
\begin{bmatrix}
2&1\\
1&1
\end{bmatrix}.
$$

The change-of-basis matrix from \(B\) to \(C\) is

$$
P_{C\leftarrow B} =
P_C^{-1}P_B.
$$

Since

$$
P_C^{-1} =
\begin{bmatrix}
1&-1\\
-1&2
\end{bmatrix},
$$

we get

$$
P_{C\leftarrow B} =
\begin{bmatrix}
1&-1\\
-1&2
\end{bmatrix}
\begin{bmatrix}
1&1\\
1&-1
\end{bmatrix} =
\begin{bmatrix}
0&2\\
1&-3
\end{bmatrix}.
$$

Therefore

$$
[v]_C=
\begin{bmatrix}
0&2\\
1&-3
\end{bmatrix}
[v]_B.
$$

If

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

then

$$
[v]_C=
\begin{bmatrix}
0&2\\
1&-3
\end{bmatrix}
\begin{bmatrix}
4\\
1
\end{bmatrix} =
\begin{bmatrix}
2\\
1
\end{bmatrix}.
$$

Check the vector itself:

$$
v=4
\begin{bmatrix}
1\\
1
\end{bmatrix}
+
1
\begin{bmatrix}
1\\
-1
\end{bmatrix} =
\begin{bmatrix}
5\\
3
\end{bmatrix}.
$$

Also,

$$
2
\begin{bmatrix}
2\\
1
\end{bmatrix}
+
1
\begin{bmatrix}
1\\
1
\end{bmatrix} =
\begin{bmatrix}
5\\
3
\end{bmatrix}.
$$

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 \(v\) remains the same. The coordinate column changes from \([v]_B\) to \([v]_C\).

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

| Matrix use | Meaning |
|---|---|
| Transformation matrix | Sends one vector to another vector |
| Change-of-basis matrix | Sends 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 \(P_B\) and \(P_C\) assumes that vectors are already written in standard coordinates, as in \(\mathbb{R}^n\). For abstract vector spaces, the column interpretation still works.

Let

$$
B=(b_1,\ldots,b_n)
$$

and

$$
C=(c_1,\ldots,c_n)
$$

be bases of \(V\).

To build \(P_{C\leftarrow B}\), write each \(b_j\) in the \(C\)-basis:

$$
b_j=a_{1j}c_1+\cdots+a_{nj}c_n.
$$

Then the \(j\)-th column of \(P_{C\leftarrow B}\) is

$$
[b_j]_C=
\begin{bmatrix}
a_{1j}\\
\vdots\\
a_{nj}
\end{bmatrix}.
$$

Thus

$$
P_{C\leftarrow B} =
(a_{ij}).
$$

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

## 24.12 Polynomial Example

Let \(P_2\) be the space of polynomials of degree at most \(2\).

Let

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

and

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

We construct \(P_{C\leftarrow B}\). Its columns are

$$
[1]_C,\qquad [x]_C,\qquad [x^2]_C.
$$

First,

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

so

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

Next,

$$
x=-1\cdot 1+1(1+x)+0(1+x+x^2),
$$

so

$$
[x]_C=
\begin{bmatrix}
-1\\
1\\
0
\end{bmatrix}.
$$

Finally,

$$
x^2=0\cdot 1-1(1+x)+1(1+x+x^2),
$$

so

$$
[x^2]_C=
\begin{bmatrix}
0\\
-1\\
1
\end{bmatrix}.
$$

Therefore

$$
P_{C\leftarrow B} =
\begin{bmatrix}
1&-1&0\\
0&1&-1\\
0&0&1
\end{bmatrix}.
$$

For

$$
p(x)=3-5x+2x^2,
$$

we have

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

Thus

$$
[p]_C =
\begin{bmatrix}
1&-1&0\\
0&1&-1\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
3\\
-5\\
2
\end{bmatrix} =
\begin{bmatrix}
8\\
-7\\
2
\end{bmatrix}.
$$

So

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

## 24.13 Matrix Example

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

$$
B=(E_{11},E_{12},E_{21},E_{22}).
$$

Let

$$
C=(A_1,A_2,A_3,A_4),
$$

where

$$
A_1=E_{11},
\quad
A_2=E_{11}+E_{12},
\quad
A_3=E_{21},
\quad
A_4=E_{21}+E_{22}.
$$

To compute coordinates in \(C\), write a matrix

$$
M=
\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}
$$

as

$$
M =
\alpha A_1+\beta A_2+\gamma A_3+\delta A_4.
$$

Then

$$
\alpha A_1+\beta A_2+\gamma A_3+\delta A_4 =
\begin{bmatrix}
\alpha+\beta&\beta\\
\gamma+\delta&\delta
\end{bmatrix}.
$$

Matching entries gives

$$
\beta=b,
\qquad
\alpha=a-b,
\qquad
\delta=d,
\qquad
\gamma=c-d.
$$

Therefore

$$
[M]_C=
\begin{bmatrix}
a-b\\
b\\
c-d\\
d
\end{bmatrix}.
$$

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

## 24.14 Change of Basis and Linear Maps

Let

$$
T:V\to V
$$

be a linear transformation. Suppose its matrix in basis \(B\) is

$$
[T]_B.
$$

If \(C\) is another basis, the matrix of the same transformation in basis \(C\) is related by a change of basis.

Let

$$
S=P_{B\leftarrow C}.
$$

This matrix converts \(C\)-coordinates into \(B\)-coordinates:

$$
[v]_B=S[v]_C.
$$

Apply the transformation in \(B\)-coordinates:

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

Convert back to \(C\)-coordinates:

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

Substitute:

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

Therefore

$$
[T]_C=S^{-1}[T]_B S.
$$

This is the similarity formula.

## 24.15 Similar Matrices

Two square matrices \(A\) and \(B\) are similar if there exists an invertible matrix \(S\) such that

$$
B=S^{-1}AS.
$$

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

If

$$
S=
\begin{bmatrix}
|&|&&|\\
v_1&v_2&\cdots&v_n\\
|&|&&|
\end{bmatrix}
$$

has eigenvectors as columns, and

$$
Av_i=\lambda_i v_i,
$$

then

$$
S^{-1}AS=
\begin{bmatrix}
\lambda_1&0&\cdots&0\\
0&\lambda_2&\cdots&0\\
\vdots&\vdots&\ddots&\vdots\\
0&0&\cdots&\lambda_n
\end{bmatrix}.
$$

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 conversion | Matrix |
|---|---|
| \(B\)-coordinates to standard coordinates | \(P_B\) |
| Standard coordinates to \(B\)-coordinates | \(P_B^{-1}\) |
| \(B\)-coordinates to \(C\)-coordinates | \(P_C^{-1}P_B\) |
| \(C\)-coordinates to \(B\)-coordinates | \(P_B^{-1}P_C\) |

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 \(\mathbb{R}^2\), 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 \(B\) to basis \(C\) in \(\mathbb{R}^n\):

| Step | Operation |
|---|---|
| 1 | Form \(P_B\) from the columns \(b_1,\ldots,b_n\) |
| 2 | Form \(P_C\) from the columns \(c_1,\ldots,c_n\) |
| 3 | Compute \(P_{C\leftarrow B}=P_C^{-1}P_B\) |
| 4 | Multiply \([v]_C=P_{C\leftarrow B}[v]_B\) |

For an abstract vector space, replace the basis matrices by coordinate columns. Express each \(B\)-basis vector in the \(C\)-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:

| Concept | Meaning |
|---|---|
| Basis matrix | Matrix whose columns are basis vectors |
| \(P_B[v]_B\) | Convert \(B\)-coordinates to standard coordinates |
| \(P_C^{-1}P_B\) | Convert \(B\)-coordinates to \(C\)-coordinates |
| \(P_{C\leftarrow B}\) | Change-of-basis matrix from \(B\) to \(C\) |
| Columns of \(P_{C\leftarrow B}\) | \(C\)-coordinates of the \(B\)-basis vectors |
| Similar matrices | Same linear operator in different bases |
| Diagonalization | Change 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.