The Complete Guide to Change of Basis in Linear Algebra with Step-by-Step Examples
This complete guide to change of basis in linear algebra will teach you how vectors, matrices, and polynomials can be represented in different coordinate systems, and how transition matrices allow us to move between them. Understanding how vectors change when we switch from one basis to another is an important idea because it provides a deeper understanding of vector spaces, coordinate representations, and linear transformations.
We begin by reviewing the idea of coordinates relative to a basis and how vectors can be expressed as linear combinations of basis vectors. From there, we introduce transition matrices, which are the main tools for converting coordinates from one basis to another. You will then learn systematic methods for constructing transition matrices and finding their inverses so you can move back and forth between coordinate systems.
By the end of this tutorial, you will be able to: compute coordinates relative to arbitrary bases, construct transition matrices step by step, convert coordinates between different bases, and invert transition matrices across multiple types of vector spaces.
Basis and Coordinate Systems
Before learning how to change from one basis to another, we first need to understand the fundamental relationship between basis and coordinate systems in vector spaces. In linear algebra, a basis provides a framework for describing vectors using coordinates, much like how Cartesian, polar, cylindrical, and spherical coordinates provide different ways to describe points in calculus.
As discussed in The Ultimate Guide to Understanding Basis and Dimension in Vector Spaces, a basis is much more than just a collection of vectors. A basis determines how every vector in a vector space is represented and interpreted. Similarly, in A Beginner-Friendly Approach to Understanding Vector Spaces and Subspaces, we saw that vector spaces can contain many different types of mathematical objects, including vectors, matrices, and polynomials. The process of changing basis is essentially changing coordinate systems within those spaces.
Coordinates of a Vector Relative to a Basis
Once a basis is selected, every vector in the vector space can be represented by a coordinate vector.
Definition
Suppose
$$B = \{ v_1, v_2, \dots, v_n \}$$
is a basis for a vector space \( V \). If a vector \( v \in V \) can be written as
$$v = c_1v_1 + c_2v_2 + \cdots + c_nv_n$$
then the coordinate vector of \( v \) relative to the basis \( B \) is
$$(v)_B = \langle c_1, c_2, \cdots, c_n \rangle.$$
The components \( c_1, c_2, \dots, c_n \) are called the coordinates of \( v \) relative to \( B \).
Coordinates Depend on the Basis
A vector does not change when we change the basis, but its coordinate representation does.
This idea is similar to switching between coordinate systems in multivariable calculus. In A Complete Beginner’s Guide to Polar, Cylindrical, and Spherical Coordinates in Calculus, different coordinate systems were used to describe the same geometric objects from different perspectives. Change of basis works similarly. The vector itself does not change, but its coordinates depend on the basis being used.
Examples
We now provide examples of finding coordinate vectors. In the process, we will use Gaussian elimination. For a review of Gaussian elimination, please refer to the article How to Solve a System of Equations Using Gaussian Elimination: A Step-by-Step Guide.
We will also make use of linear combinations. For a review of linear combinations, please refer to the articles The Ultimate Step-by-Step Guide to Basic Matrix Operations for Beginners, Vector Operations Tutorial for Beginners: A Complete Step-by-Step Guide, and Linear Dependence vs Independence Explained for Beginners: Everything You Need to Know.
Example 1: Find the coordinate vectors relative to the given basis:
(a) $$B = \{ \langle 1, 1 \rangle, \langle 1, -1 \rangle \}, v = \langle 4, 2 \rangle$$
(b) $$B = \{\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix},
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix} \}, A = \begin{bmatrix}
2 & -1 \\
5 & 3
\end{bmatrix}.$$
(c) $$B = \{1+x, x+x^2, 1+x^2 \}, p(x) = 4 + 3x + 5x^2.$$
Solution: (a) $$B = \{ \langle 1, 1 \rangle, \langle 1, -1 \rangle \}, v = \langle 4, 2 \rangle.$$
We seek constants \( c_1 \) and \( c_2 \) such that
$$c_1\langle 1, 1 \rangle + c_2\langle 1, -1 \rangle = \langle 4, 2 \rangle.$$
Multiplying each component of the first vector by \( c_1 \) and each component of the second vector by \( c_2 \) gives
$$\langle c_1, c_1 \rangle + \langle c_2, -c_2 \rangle = \langle 4, 2 \rangle.$$
Adding corresponding components gives
$$\langle c_1 + c_2, c_1 – c_2 \rangle = \langle 4, 2 \rangle.$$
Next, we equate components to obtain the system of equations
$$\begin{aligned}
c_1 + c_2 &= 4 \\
c_1 – c_2 &= 2
\end{aligned}.$$
The augmented matrix for the system is
$$\begin{pmatrix}
1 & 1 & 4 \\
1 & -1 & 2
\end{pmatrix}.$$
Subtracting 1 times row 1 from row 2, we get
$$\begin{pmatrix}
1 & 1 & 4 \\
1-1(1) & -1 – 1(1) & 2 – 1(4)
\end{pmatrix}.$$
Simplifying we get
$$\begin{pmatrix}
1 & 1 & 4 \\
1-1 & -1 – 1 & 2 – 4
\end{pmatrix}.$$
Subtracting, we arrive at
$$\begin{pmatrix}
1 & 1 & 4 \\
0 & -2 & -2
\end{pmatrix}.$$
Dividing row 2 by -2 gives
$$\begin{pmatrix}
1 & 1 & 4 \\
\frac{0}{-2} & \frac{-2}{-2} & \frac{-2}{-2}
\end{pmatrix}.$$
Dividing, we obtain
$$\begin{pmatrix}
1 & 1 & 4 \\
0 & 1 & 1
\end{pmatrix}.$$
In equation form, this is
$$\begin{aligned}
c_1 + c_2 &= 4 \\
c_2 &= 1
\end{aligned}.$$
Substituting \( c_2 = 1 \) into the first equation gives
$$\begin{aligned}
c_1 + 1 &= 4 \\
c_2 &= 1
\end{aligned}.$$
Which implies
$$\begin{aligned}
c_1 &= 3 \\
c_2 &= 1
\end{aligned}.$$
Thus, our solution is
$$c_1 = 3, c_2 = 1.$$
Plugging into the original equation, then gives
$$3\langle 1, 1 \rangle + 1\langle 1, -1 \rangle = \langle 4, 2 \rangle.$$
Hence, the final coordinate vector is
$$(v)_B = \langle 3, 1 \rangle.$$
$$B = \{\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix},
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}\}, A = \begin{bmatrix}
2 & -1 \\
5 & 3
\end{bmatrix}.$$
Notice that
$$2\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}
– 1\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
+ 5\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}
+ 3\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix} = \begin{bmatrix}
2 & -1 \\
5 & 3
\end{bmatrix}.$$
Hence, the final coordinate vector is
$$(A)_B = \langle 2, -1, 5, 3 \rangle.$$
(c) $$B = \{ 1+x, x+x^2, 1+x^2 \}, p(x) = 4 + 3x + 5x^2.$$
We seek constants \( c_1 \), \( c_2 \), and \( c_3 \) such that
$$c_1(1 + x) + c_2(x + x^2) + c_3(1 + x^2) = 4 + 3x + 5x^2.$$
Expanding the left-hand side of the equation, we obtain
$$c_1 + c_1x + c_2x + c_2x^2 + c_3 + c_3x^2 = 4 + 3x + 5x^2.$$
Adding like terms, we find
$$(c_1 + c_3) + (c_1 + c_2)x + (c_2+c_3)x^2 = 4 + 3x + 5x^2.$$
Next, we equate coefficients to obtain the system of equations
$$\begin{aligned}
c_1 + c_3 &= 4 \\
c_1 + c_2 &= 3 \\
c_2 + c_3 &= 5
\end{aligned}.$$
The augmented matrix for the system is
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
1 & 1 & 0 & 3 \\
0 & 1 & 1 & 5
\end{pmatrix}.$$
Subtracting 1 times row 1 from row 2 we get
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
1 – 1(1) & 1 – 1(0) & 0 – 1(1) & 3 – 1(4) \\
0 & 1 & 1 & 5
\end{pmatrix}.$$
Simplifying we get
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
1 – 1 & 1 – 0 & 0 – 1 & 3 – 4 \\
0 & 1 & 1 & 5
\end{pmatrix}.$$
Subtracting, we arrive at
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
0 & 1 & –1 & -1 \\
0 & 1 & 1 & 5
\end{pmatrix}.$$
Subtracting 1 times row 2 from row 3 we get
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
0 & 1 & –1 & -1 \\
0 – 1(0) & 1 – 1(1) & 1 – 1(-1) & 5 – 1(-1)
\end{pmatrix}.$$
Simplifying we get
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
0 & 1 & –1 & -1 \\
0 – 0 & 1 – 1 & 1 + 1 & 5 + 1
\end{pmatrix}.$$
Subtracting and adding, we arrive at
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
0 & 1 & –1 & -1 \\
0 & 0 & 2 & 6
\end{pmatrix}.$$
Dividing row 3 by 2 gives
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
0 & 1 & –1 & -1 \\
\frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{6}{2}
\end{pmatrix}.$$
Dividing, we obtain
$$\begin{pmatrix}
1 & 0 & 1 & 4 \\
0 & 1 & –1 & -1 \\
0 & 0 & 1 & 3
\end{pmatrix}.$$
In equation form, this is
$$\begin{aligned}
c_1 + c_3 &= 4 \\
c_2 – c_3 &= -1 \\
c_3 &= 3
\end{aligned}.$$
Substituting \( c_3 = 3 \) into the first and second equations gives
$$\begin{aligned}
c_1 + 3 &= 4 \\
c_2 – 3 &= -1 \\
c_3 &= 3
\end{aligned}.$$
Which implies
$$\begin{aligned}
c_1 &= 1 \\
c_2 &= 2 \\
c_3 &= 3
\end{aligned}.$$
Thus, our solution is
$$c_1 = 1, c_2 = 2, c_3 = 3.$$
Plugging into the original equation, then gives
$$1(1 + x) + 2(x + x^2) + 3(1 + x^2) = 4 + 3x + 5x^2.$$
Hence, the final coordinate vector is
$$(p(x))_B = \langle 1, 2, 3 \rangle.$$
Transition Matrices and Change of Basis
In the previous 2 sections, we learned that coordinates depend on the basis chosen for a vector space. The next step is learning how to move between coordinate systems. This process is called a change of basis, and the main tool for performing it is the transition matrix.
Transition matrices play a role in linear algebra similar to that of coordinate transformations in multivariable calculus. In Beginner’s Guide to Changing Variables in Double and Triple Integrals: Everything You Need to Know, changing variables allowed us to rewrite problems in more convenient coordinate systems. The underlying geometric object remained the same, but the coordinates describing it changed. Change of basis works in the same way for vector spaces.
There is also a strong connection between transition matrices and linear transformations. As discussed in Step-by-Step Solutions to Common Examples of Linear Transformations in Linear Algebra, matrices are used to describe how vectors transform under mappings between vector spaces. A transition matrix can be viewed as a special type of linear transformation that converts coordinates from one basis to another.
Definition
Suppose:
$$B = \{ v_1, v_2, \dots, v_n \}$$
and
$$C = \{ w_1 , w_2, \dots, w_n \}$$
are two bases for a vector space \( V \).
$$P$$
is the transition matrix from basis \( B \) to basis \( C \), whose columns consist of the coordinate vectors of B relative to the basis C.
Step-by-Step Process for Finding Transition Matrices
To find a transition matrix between two bases:
Step 1: Identify the Old and New Bases
Determine the starting basis and the target basis
Step 2: Express Basis Vectors in Terms of the Target Basis
Rewrite each vector from the original basis as a linear combination of the target basis vectors.
Step 3: Form Coordinate Vectors as Columns
The coefficients from each coordinate vector become the columns of the transition matrix.
Step 4: Assemble the Transition Matrix
Arrange the coordinate vectors into a matrix.
Step 5: Multiply by a Coordinate Vector from the Old Basis
This gives the coordinates relative to the new basis.
Change of Basis Examples
Examples in \( \mathbb{R}^n \)
Example 2: Find the transition matrices from B to C, then use them to convert \( (v)_B \) to \( (v)_C \):
(a) $$B = \{ \langle 1, 1 \rangle, \langle 1, -1 \rangle \}, C = \{ \langle 1, 0 \rangle, \langle 0, 1 \rangle \}, (v)_B = \langle 3, 1 \rangle$$
(b) $$B = \{ \langle 1, 0, 1 \rangle, \langle 0, 1, 1 \rangle, \langle 1, 1, 0 \}, C = \{ \langle 1, 0, 0 \rangle, \langle 0, 1, 0 \rangle, \langle 0, 0, 1 \}, (v)_B = \langle 2, -1, 3 \rangle.$$
Solution: (a) $$B = \{ \langle 1, 1 \rangle, \langle 1, -1 \rangle \}, C = \{ \langle 1, 0 \rangle, \langle 0, 1 \rangle \}, (v)_B = \langle 3, 1 \rangle.$$
Notice that
$$\langle 1, 1 \rangle = 1\langle 1, 0 \rangle + 1\langle 0, 1 \rangle.$$
The first coordinate vector is then
$$(b_1)_C = \langle 1, 1 \rangle.$$
As a column vector, this is
$$(b_1)_C = \begin{bmatrix}
1 \\
1
\end{bmatrix}.$$
Also, notice that
$$\langle 1, -1 \rangle = 1\langle 1, 0 \rangle – 1\langle 0, 1 \rangle.$$
The second coordinate vector is then
$$(b_2)_C = \langle 1, -1 \rangle.$$
As a column vector, this is
$$(b_2)_C = \begin{bmatrix}
1 \\
-1
\end{bmatrix}.$$
Forming the transition matrix, we get
$$P = \begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.$$
As a column vector \( (v)_B \) is
$$(v)_B = \begin{bmatrix}
3 \\
1
\end{bmatrix}.$$
\( (v)_C \) is then
$$(v)_C = \begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}\begin{bmatrix}
3 \\
1
\end{bmatrix}.$$
Multiplying each row of the first matrix by each column of the second matrix, we get
$$(v)_C = \begin{bmatrix}
1(3) + 1(1) \\
1(3) – 1(1)
\end{bmatrix}.$$
Multiplying we obtain
$$(v)_C = \begin{bmatrix}
3 + 1 \\
3 – 1
\end{bmatrix}.$$
Adding and subtracting, we arrive at
$$(v)_C = \begin{bmatrix}
4 \\
2
\end{bmatrix}.$$
Converting to a vector, we arrive at a final answer of
$$\langle 4, 2 \rangle.$$
(b) $$B = \{ \langle 1, 0, 1 \rangle, \langle 0, 1, 1 \rangle, \langle 1, 1, 0 \}, C = \{ \langle 1, 0, 0 \rangle, \langle 0, 1, 0 \rangle, \langle 0, 0, 1 \}, (v)_B = \langle 2, -1, 3 \rangle.$$
Notice that
$$\langle 1, 0, 1 \rangle = 1\langle 1, 0, 0 \rangle + 0\langle 0, 1, 0 \rangle + 1\langle 0, 0, 1 \rangle.$$
The first coordinate vector is then
$$(b_1)_C = \langle 1, 0, 1 \rangle.$$
As a column vector, this is
$$(b_1)_C = \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}.$$
Also, notice that
$$\langle 0, 1, 1 \rangle = 0\langle 1, 0, 0 \rangle + 1\langle 0, 1, 0 \rangle + 1\langle 0, 0, 1 \rangle.$$
The second coordinate vector is then
$$(b_2)_C = \langle 0, 1, 1 \rangle.$$
As a column vector, this is
$$(b_2)_C = \begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}.$$
Further, notice that
$$\langle 1, 1, 0 \rangle = 1\langle 1, 0, 0 \rangle + 1\langle 0, 1, 0 \rangle + 0\langle 0, 0, 1 \rangle.$$
The third coordinate vector is then
$$(b_3)_C = \langle 1, 1, 0 \rangle.$$
As a column vector, this is
$$(b_3)_C = \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}.$$
Forming the transition matrix, we get
$$P = \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{bmatrix}.$$
As a column vector \( (v)_B \) is
$$(v)_B = \begin{bmatrix}
2 \\
-1 \\
3
\end{bmatrix}.$$
\( (v)_C \) is then
$$(v)_C = \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{bmatrix}\begin{bmatrix}
2 \\
-1 \\
3
\end{bmatrix}.$$
Multiplying each row of the first matrix by each column of the second matrix, we get
$$(v)_C = \begin{bmatrix}
1(2) + 0(-1) + 1(3) \\
0(2) + 1(-1) + 1(3) \\
1(2) + 1(-1) + 0(3)
\end{bmatrix}.$$
Multiplying we obtain
$$(v)_C = \begin{bmatrix}
2 + 0 + 3 \\
0 – 1 + 3 \\
2 – 1 + 0
\end{bmatrix}.$$
Adding and subtracting, we arrive at
$$(v)_C = \begin{bmatrix}
5 \\
2 \\
1
\end{bmatrix}.$$
Converting to a vector, we arrive at a final answer of
$$\langle 5, 2, 1 \rangle.$$
Example with Matrices
Example 3: Find the transition matrix from B to C, then use it to convert \( (v)_B \) to \( (v)_C \):
$$ B = \{\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}, \begin{bmatrix}
0 & 0 \\
1 & 1
\end{bmatrix}, \begin{bmatrix}
1 & -1 \\
0 & 0
\end{bmatrix}, \begin{bmatrix}
0 & 0 \\
1 & -1
\end{bmatrix}\}, C = \{ \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}, \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}, \begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}, \begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix} \}, (v)_B = \begin{bmatrix}
1 & 2 \\
2 & 4
\end{bmatrix}.$$
Solution: Notice that
$$\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix} = 1\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} + 1\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} + 0\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix} + 0\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}.$$
The first coordinate vector is then
$$(b_1)_C = \langle 1, 1, 0, 0 \rangle.$$
As a column vector, this is
$$(b_1)_C = \begin{bmatrix}
1 \\
1 \\
0 \\
0
\end{bmatrix}.$$
Also, notice that
$$\begin{bmatrix}
0 & 0 \\
1 & 1
\end{bmatrix} = 0\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} + 0\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} + 1\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix} + 1\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}.$$
The second coordinate vector is then
$$(b_2)_C = \langle 0, 0, 1, 1 \rangle.$$
As a column vector, this is
$$(b_2)_C = \begin{bmatrix}
0 \\
0 \\
1 \\
1
\end{bmatrix}.$$
Further, notice that
$$\begin{bmatrix}
1 & -1 \\
0 & 0
\end{bmatrix} = 1\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} – 1\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} + 0\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix} + 0\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}.$$
The third coordinate vector is then
$$(b_3)_C = \langle 1, -1, 0, 0 \rangle.$$
As a column vector, this is
$$(b_3)_C = \begin{bmatrix}
1 \\
-1 \\
0 \\
0
\end{bmatrix}.$$
Finally, notice that
$$\begin{bmatrix}
0 & 0 \\
1 & -1
\end{bmatrix} = 0\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} + 0\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} + 1\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix} – 1\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}.$$
The fourth coordinate vector is then
$$(b_4)_C = \langle 0, 0, 1, -1 \rangle.$$
As a column vector, this is
$$(b_4)_C = \begin{bmatrix}
0 \\
0 \\
1 \\
-1
\end{bmatrix}.$$
Forming the transition matrix, we get
$$P = \begin{bmatrix}
1& 0 & 1 & 0 \\
1& 0 & -1 & 0 \\
0 & 1 & 0 &1 \\
0 & 1 & 0 & -1
\end{bmatrix}.$$
As a column vector \( (v)_B \) is
$$(v)_B = \begin{bmatrix}
1 \\
2 \\
2 \\
4
\end{bmatrix}.$$
\( (v)_C \) is then
$$(v)_C = \begin{bmatrix}
1& 0 & 1 & 0 \\
1& 0 & -1 & 0 \\
0 & 1 & 0 &1 \\
0 & 1 & 0 & -1
\end{bmatrix}\begin{bmatrix}
1 \\
2 \\
2 \\
4
\end{bmatrix}.$$
Multiplying each row of the first matrix by each column of the second matrix, we get
$$(v)_C = \begin{bmatrix}
1(1) + 0(2) + 1(2) + 0(4) \\
1(1) + 0(2) – 1(2) + 0(4) \\
0(1) + 1(2) + 0(2) + 1(4) \\
0(1) + 1(2) + 0(2) – 1(4)
\end{bmatrix}.$$
Multiplying we obtain
$$(v)_C = \begin{bmatrix}
1 + 0 + 2 + 0 \\
1 + 0 – 2 + 0 \\
0 + 2 + 0 + 4 \\
0 + 2 + 0 – 4
\end{bmatrix}.$$
Adding and subtracting, we arrive at
$$(v)_C = \begin{bmatrix}
3 \\
-1 \\
6 \\
-2
\end{bmatrix}.$$
Converting to a matrix, we arrive at a final answer of
$$(v)_C = \begin{bmatrix}
3 & -1 \\
6 & -2
\end{bmatrix}.$$
Example with Polynomials
Example 4: Find the transition matrix from B to C, then use it to convert \( (v)_B \) to \( (v)_C \): $$B = \{ 1+x, x+x^2, 1+x^2 \}, C = \{ 1, x, x^2 \}, (v)_B = 1 + x + x^2.$$
Solution: Notice that
$$1 + x = 1(1) + 1x + 0x^2.$$
The first coordinate vector is then
$$(b_1)_C = \langle 1, 1, 0 \rangle.$$
As a column vector, this is
$$(b_1)_C = \begin{bmatrix}
1 \\
1\\\
0
\end{bmatrix}.$$
Also, notice that
$$x + x^2 = 0(1) + 1x + 1x^2.$$
The second coordinate vector is then
$$(b_2)_C = \langle 0, 1, 1 \rangle.$$
As a column vector, this is
$$(b_2)_C = \begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}.$$
Further, notice that
$$1 + x^2 = 1(1) + 0x + 1x^2.$$
The third coordinate vector is then
$$(b_3)_C = \langle 1, 0, 1 \rangle.$$
As a column vector, this is
$$(b_3)_C = \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}.$$
Forming the transition matrix, we get
$$P = \begin{bmatrix}
1 & 0 &1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{bmatrix}.$$
As a column vector \( (v)_B \) is
$$(v)_B = \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}.$$
\( (v)_C \) is then
$$(v)_C = \begin{bmatrix}
1 & 0 &1 \\
1 & 1 & 0 \\
0 & 1 & 1
\end{bmatrix}\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}.$$
Multiplying each row of the first matrix by each column of the second matrix, we get
$$(v)_C = \begin{bmatrix}
1(1) + 0(1) + 1(1) \\
1(1) + 1(1) + 0(1) \\
0(1) + 1(1) + 1(1)
\end{bmatrix}.$$
Multiplying we obtain
$$(v)_C = \begin{bmatrix}
1 + 0 + 1 \\
1 + 1 + 0 \\
0 + 1 + 1
\end{bmatrix}.$$
Adding, we arrive at
$$(v)_C = \begin{bmatrix}
2 \\
2 \\
2
\end{bmatrix}.$$
Converting to a polynomial, we arrive at a final answer of
$$(v)_C = 2 + 2x + 2x^2.$$
Inverse Transition Matrices
In the last 3 sections, we learned how transition matrices convert coordinates from one basis to another. However, in many applications, we also need to reverse the process and return to the original coordinate system. This is where inverse transition matrices are needed.
An inverse transition matrix allows us to move backward between bases. If a transition matrix converts coordinates from basis \( B \) to basis \( C \), then its inverse converts coordinates from basis \( C \) back to basis \( B \).
The process of finding inverse transition matrices relies heavily on matrix inversion techniques, as discussed in the article How to Find the Inverse of a Matrix Step by Step: A Complete Beginner’s Guide.
Example
Example 5: P is a transition matrix from B to C, find \( (v)_B \):
$$P = \begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}, (v)_C = \begin{bmatrix}
4 \\
2
\end{bmatrix}.$$
Solution: Applying the formula for \( 2 \times 2 \) inverse matrices gives
$$P^{-1} = \frac{1}{1(-1) – 1(1)}\begin{bmatrix}
-1 & -1 \\
-1 & 1
\end{bmatrix}.$$
Multiplying we obtain
$$P^{-1} = \frac{1}{-1 – 1}\begin{bmatrix}
-1 & -1 \\
-1 & 1
\end{bmatrix}.$$
Subtracting we get
$$P^{-1} = \frac{1}{-2}\begin{bmatrix}
-1 & -1 \\
-1 & 1
\end{bmatrix}.$$
Dividing gives
$$P^{-1} = -\frac{1}{2}\begin{bmatrix}
-1 & -1 \\
-1 & 1
\end{bmatrix}.$$
Multiplying each entry by \( -\frac{1}{2} \) we obtain
$$P^{-1} = \begin{bmatrix}
\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2}
\end{bmatrix}.$$
\( (v)_B \) is then
$$(v)_B = \begin{bmatrix}
\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2}
\end{bmatrix}\begin{bmatrix}
4 \\
2
\end{bmatrix}.$$
Multiplying each row of the first matrix by each column of the second matrix, we get
$$(v)_B = \begin{bmatrix}
\frac{1}{2}(4) + \frac{1}{2}(2) \\
\frac{1}{2}(4) – \frac{1}{2}(2)
\end{bmatrix}.$$
Multiplying we obtain
$$(v)_B = \begin{bmatrix}
2 + 1 \\
2 – 1
\end{bmatrix}.$$
Adding and subtracting, we arrive at a final answer of
$$(v)_B = \begin{bmatrix}
3 \\
1
\end{bmatrix}.$$
Conclusion
In this complete guide to change of basis in linear algebra, we explored how different bases provide different coordinate systems for describing the same vectors. We saw that coordinates depend entirely on the chosen basis and that transition matrices provide a systematic way to move between coordinate systems. By learning how to construct transition matrices step by step, we gained the ability to convert coordinates in spaces such as \( \mathbb{R}^n \), matrix vector spaces, and polynomial vector spaces. We also studied inverse transition matrices and learned how inverse matrices allow us to reverse basis transformations and recover coordinates in the original basis.
Further Reading
A Comprehensive Beginner’s Guide to Partial Fraction Decomposition – With your knowledge of coordinate vectors, I recommend revisiting partial fraction decomposition and noting the similarities between the two topics.
