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Change of basis

by , PhD

The change of basis is a technique that allows us to express vector coordinates with respect to a "new basis" that is different from the "old basis" originally employed to compute coordinates.

Table of Contents

Coordinates

Suppose that a finite-dimensional vector space $S$ possesses a basis [eq1].

Then, any vector $sin S$ can be written as a linear combination of the basis:[eq2]where the scalar coefficients [eq3] are uniquely determined.

Remember that the Kx1 vector[eq4]is called the coordinate vector of $s$ with respect to the basis $B$.

Example Let $S$ be a vector space and [eq5] a basis for $S $. Consider the vector[eq6]Its coordinate vector is[eq7]

The change-of-basis matrix

Suppose that we have a second basis [eq8]. By the dimension theorem, $B$ and $C$ have the same number K of vectors.

The first question we are going to ask is: what happens to coordinates when we switch from using $B$ as a basis to using $C$? In particular, how do we transform a coordinate vector [eq9] into a vector [eq10] of coordinates with respect to the new basis?

The answer is provided by the following proposition.

Proposition Let $S$ be a vector space. Let [eq11] and [eq12] be two bases for $S$. Then, there exists a $K	imes K$ matrix, denoted by $S_{B
ightarrow C}$ and called change-of-basis matrix, such that, for any $sin S$,[eq13]where [eq14] and [eq15] denote the coordinate vectors of $s$ with respect to $B$ and $C$ respectively.

Proof

Let [eq16]be the representation of $s$ in terms of $B$. By the rules on the addition and scalar multiplication of coordinate vectors, we have that[eq17]where [eq18] are the Kx1 coordinate vectors of the elements of $B$ with respect to $C$. Adjoin these K vectors so as to form a $K	imes K$ matrix[eq19]Since[eq20]we can write equation (1) as[eq21]because the product [eq22] is equal to a linear combination of the columns of $S_{B
ightarrow C}$, with coefficients taken from [eq14] (see the lecture on matrix products and linear combinations). Note that $S_{B
ightarrow C}$ does not depend on the particular choice of $s$, as it depends only on the two bases $B$ and $C$.

The main take-away from the previous proof is that the columns of the change-of-basis matrix $S_{B
ightarrow C}$ are the coordinates of the vectors of the original basis $B$ with respect to the new basis $C$:[eq24]

As demonstrated by the next proposition, the change of basis matrix is invertible.

Proposition Let $S$ be a vector space. Let [eq11] and [eq12] be two bases for $S$. Then, the change-of-basis matrix $S_{B
ightarrow C}$ is invertible and its inverse equals $S_{C
ightarrow B}$, that is,[eq27]

Proof

For any $sin S$, we have that[eq21]and[eq29]By combining these two equations, we obtain[eq30]This can be true for every $sin S$ only if [eq31]where I is the $K	imes K$ identity matrix. The latter result implies that $S_{C
ightarrow B}$ is the inverse of $S_{B
ightarrow C}$.

Let us make an example.

Example Consider the space $S$ of all $2	imes 1$ vectors and the two bases [eq32] with[eq33]and [eq34] with[eq35]We have[eq36]Thus, the coordinate vectors of the elements of $B$ with respect to $C$ are [eq37]Therefore, when we switch from $B$ to $C$, the change-of-basis matrix is[eq38]For example, take a vector[eq39]Since[eq40]the coordinates of $s$ with respect to $B$ are[eq41]Its coordinates with respect to $C$ can be easily computed thanks to the change-of-basis matrix:[eq42]We can easily check that this is correct:[eq43]

Effect on the matrix of a linear operator

Remember that a linear operator on a vector space $S$ is a function $f:S
ightarrow S$ such that[eq44]for any two vectors $s_{1},s_{2}in S$ and any two scalars $lpha _{1}$ and $lpha _{2}$.

Given a basis [eq45] for $S$, the matrix of the linear operator with respect to $B$ is the square $K	imes K$ matrix [eq46] such that[eq47]for any vector $sin S$ (see also the lecture on the matrix of a linear map). In other words, if you multiply the matrix of the operator by the coordinate vector of $s$, then you obtain the coordinate vector of $fleft( s
ight) $.

What happens to the matrix of the operator when we switch to a new basis? The next proposition provides an answer to this question.

Proposition Let $S$ be a linear space. Let [eq48] and [eq12] be two bases for $S$. Let $f:S
ightarrow S$ be a linear operator. Denote by [eq50] and [eq51] the matrices of the linear operator with respect to $B $ and $C$ respectively. Then, [eq52]or, equivalently,[eq53]where $S_{B
ightarrow C}$ and $S_{C
ightarrow B}$ are the change-of-basis matrices that allow to switch from $B$ to $C$ and vice versa.

Proof

Let $sin S$. We can use the change-of-basis matrix $S_{C
ightarrow B}$ to transform the coordinates[eq29] and [eq55]Therefore, the matrix representation of the operator[eq47]can be written as[eq57]or[eq58]Thus, the matrix[eq59]is the matrix of $f$ with respect to $C$ (which is unique). In other words,[eq60]Since [eq61]we can also write[eq53]

Thus, the change-of-basis matrices allow to easily switch from the matrix of the linear operator with respect to the old basis to the matrix with respect to the new basis.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let us consider the space $S$ introduced in the example above with the two bases [eq63] and [eq34]. In that example, we have shown that the change-of-basis matrix is[eq65]Moreover,[eq66]Let $f$ be the linear operator such that[eq67]Find the matrix [eq68] and then use the change-of-basis formulae to derive [eq51] from [eq68].

Solution

The matrix of the linear operator with respect to the basis $B$ is[eq71]The change-of-basis formula gives[eq72]

How to cite

Please cite as:

Taboga, Marco (2017). "Change of basis", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/change-of-basis.

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