Serlo: EN: Basis change via matrices

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In this article, you will learn about basis change via matrices. Basis change matrices can be used to convert coordinates with respect to a given basis into coordinates with respect to another basis. This is particularly useful for matrices of linear maps, which are always taken with respect to two specific bases.

Derivation

We have seen in the article on bases that every finite-dimensional vector space has a basis. This means if $V$ is an $n$-dimensional $K$-vector space, then there is a basis $B=\{b_1,\dots ,b_n\}$ of $V$. Every vector $v\in V$ can therefore be written uniquely as a linear combination of the basis vectors $b_1,\dots ,b_n$, i.e. $v=\sum _{i=1}^n{\lambda }_i{b}_i$ with unique ${\lambda }_1,\dots ,{\lambda }_n\in K$.

We also know that vector spaces usually have more than one basis. Let $C=\{c_1,\dots ,c_n\}$ be a second basis of $V$. Then we can also write $v$ uniquely as a linear combination of $c_i$, i.e. $v=\sum _{i=1}^n{\mu }_i{c}_i$ with unique coefficients ${\mu }_1,\dots ,{\mu }_n\in K$.

We therefore have two representations of the vector $v$. Using the basis $B$ we get the representation $v=\sum _{i=1}^n{\lambda }_i{b}_i$ and using the basis $C$ we get $v=\sum _{i=1}^n{\mu }_i{c}_i$.

How can we convert the basis representation with respect to $B$ of the vector $v$ into the representation with respect to $C$?

This question is particularly interesting in the context of matrices of linear maps, as we will see below in the section Application of basis change via matrices. Mapping matrices allow us to calculate with coordinates instead of vectors of $V$. However, the coordinates of a vector always depend on the chosen basis in $V$. We want a simple way to convert the coordinates of any vector in $V$ with respect to a basis $B$ into coordinates with respect to another basis $C$.

The situation in $K^n$

To answer this question, we start with a simpler special case. We consider $K^n$ as a vector space and set $B=(e_1,\dots ,e_n)$ as the (ordered) standard basis. Let further $C=(c_1,\dots ,c_n)$ be any ordered basis of $K^n$. Since matrices of linear maps depend on the order of the basis vectors, we have to use ordered bases $B$ and $C$.

Let $v=(x_1,\dots ,x_n{)}^T=\sum _{i=1}^n{x}_i{e}_i$ be a vector for whom we know the coordinates with respect to the standard basis $B$. The vector $v\in K^n$ can be written in the basis $C$ as $v={\lambda }_1{c}_1+\cdots +{\lambda }_n{c}_n$ for unique ${\lambda }_1,\dots ,{\lambda }_n\in K$. How can we calculate the coordinates ${\lambda }_1,\dots ,{\lambda }_n\in K$ of $v$ with respect to $C$ simply from the coordinates $x_1,\dots ,x_n$ of $v$ with respect to the standard basis $B$?

To do this, we need to describe the mapping $K^n\to K^n$, which maps each vector $v=(x_1,...,x_n{)}^T\in K^n$ to its coordinate vector $({\lambda }_1,\dots ,{\lambda }_n{)}^T\in K^n$ with respect to $C$. This is done by the coordinate mapping $k_C:K^n\to K^n$, which is a linear map that we know from the article on isomorphims.

In order to describe $k_C$, we calculate its matrix $M_{\mathrm{S}\mathrm{t}\mathrm{d}}^{\mathrm{S}\mathrm{t}\mathrm{d}}(k_C)$ with respect to the standard basis $B=(e_1,\dots ,e_n)$. By using matrix-vector multiplication in $K^n$, we then obtain the coordinate vector $({\lambda }_1,\dots ,{\lambda }_n{)}^T$ by multiplying $v=(x_1,\dots ,x_n{)}^T$ from the left by $M_{\mathrm{S}\mathrm{t}\mathrm{d}}^{\mathrm{S}\mathrm{t}\mathrm{d}}(k_C)$.

To calculate the matrix $M_{\mathrm{S}\mathrm{t}\mathrm{d}}^{\mathrm{S}\mathrm{t}\mathrm{d}}(k_C)$, we need to determine $k_C(e_1),\dots ,k_C(e_n)$. These will then be the columns of $M_{\mathrm{S}\mathrm{t}\mathrm{d}}^{\mathrm{S}\mathrm{t}\mathrm{d}}(k_C)$. We are therefore looking for the coordinates of $e_1,\dots ,e_n$ with respect to $C$, so we have to write these as a linear combination of vectors in $C$. This gives us $n$ equations Vorlage:Einrücken where $a_{ij}$ are the coordinates we are looking for. The coefficients $a_{ij}$ can be determined by solving a linear system of equations. Mathe für Nicht-Freaks: Vorlage:Beispiel

Then $k_C(e_j)=(a_{1j},a_{2j},\dots ,a_{nj}{)}^T$ for $j=1,\dots ,n$. This gives us the matrix Vorlage:Einrücken We obtain $M_{\mathrm{S}\mathrm{t}\mathrm{d}}^{\mathrm{S}\mathrm{t}\mathrm{d}}(k_C)y=k_C(y)$ for all $y\in K^n$. The required coefficients ${\lambda }_1,\dots ,{\lambda }_n$ are therefore obtained by Vorlage:Einrücken

Generalization to arbitrary finite-dimensional vector spaces

In a general finite-dimensional vector space $V$, unlike in $K^n$, there is no standard basis. In this situation, we have two ordered bases $B=(b_1,\dots ,b_n)$ and $C=(c_1,\dots ,c_n)$. Usually, we are then given an arbitrary vector $v\in V$ as a linear combination $v=x_1{b}_1+\dots +x_n{b}_n$ with respect to the basis $B$ with $x_1,\dots ,x_n\in K$. The coefficients $x_1,\dots ,x_n$ are also called the coordinates of $v$ with respect to $B$. Correspondingly, the coordinates with respect to $C$ are the scalars ${\lambda }_1,\dots ,{\lambda }_n\in K$ with $v={\lambda }_1{c}_1+\dots +{\lambda }_n{c}_n$.

We are looking for a method to convert the coordinates $x_1,\dots ,x_n$ with respect to $B$ of any vector $v\in V$ into the coordinates ${\lambda }_1,\dots ,{\lambda }_n$ with respect to $C$. For this, we need a mapping $K^n\to K^n$, which sends $(x_1,\dots ,x_n{)}^T$ to $({\lambda }_1,\dots ,{\lambda }_n{)}^T$.

We already know the coordinate mappings $k_B:V\to K^n$ with $k_B(v)=(x_1,\dots ,x_n{)}^T\in K^n$ and $k_C:V\to K^n$ with $k_C(v)=({\lambda }_1,\dots ,{\lambda }_n{)}^T$. From $(x_1,\dots ,x_n{)}^T\in K^n$ we want to obtain the vector $({\lambda }_1,\dots ,{\lambda }_n{)}^T\in K^n$. The coordinate mappings are isomorphisms. So $k_B^{-1}:K^n\to V$ maps the vector $(x_1,\dots ,x_n{)}^T$ to $v$ and $k_C:V\to K^n$ maps $v$ to $({\lambda }_1,\dots ,{\lambda }_n{)}^T$. If we first execute $k_B^{-1}$ and then $k_C$, we obtain a mapping that sends $(x_1,\dots ,x_n{)}^T$ to $({\lambda }_1,\dots ,{\lambda }_n{)}^T$.

Vorlage:Anker Our desired transformation is therefore realized by the linear map $k_C\circ k_B^{-1}:K^n\to K^n$. As above for the situation in $K^n$, we can then determine the matrix of this linear map in $K^n$ with respect to the standard basis. This matrix is given by $M_{\mathrm{S}\mathrm{t}\mathrm{d}}^{\mathrm{S}\mathrm{t}\mathrm{d}}(k_C\circ k_B^{-1})$. If we remember the article on matrices of linear maps, however, this matrix is just $M_C^B({\mathrm{id}}_V)$, because $k_C\circ k_B^{-1}=k_C\circ {\mathrm{id}}_V\circ k_B^{-1}$.

It also makes intuitive sense that the matrix executing the basis change from $B$ to $C$ is given exactly by $M_C^B({\mathrm{id}}_V)$ representing the identity from basis $B$ to $C$. This is because, if we multiply the coordinate vector $k_B(v)$ of $v\in V$ with respect to $B$ from the left with $M_C^B({\mathrm{id}}_V)$, then we obtain exactly the coordinate vector of ${\mathrm{id}}_V(v)=v$ with respect to $C$, just by definition of the representing matrix. That is, Vorlage:Einrücken for all $v\in V$. The matrix $M_C^B({\mathrm{id}}_V)$ therefore converts coordinates with respect to $B$ into coordinates with respect to $C$. This is exactly what a basis change matrix does.

Definition

Mathe für Nicht-Freaks: Vorlage:Definition The basis change matrix has many other names. It is also referred to in the literature as a transition matrix, basis transition matrix, transformation matrix or coordinate change matrix. Mathe für Nicht-Freaks: Vorlage:Warnung

Application of basis change via matrices

The problem with matrices of linear maps

We can find a matrix $M_C^B(f)$ for every linear map $f:V\to W$ between two finite-dimensional vector spaces, with respect to bases $B$ and $C$. However, this matrix depends on $B$ and $C$, and their order. If we choose other bases $B^{\prime }$ or $C^{\prime }$, we will very likely get a different matrix. We can see this in the following example: Mathe für Nicht-Freaks: Vorlage:Beispiel

Solution of this problem

Consider a linear map $f:V\to W$ and two ordered bases $B$ and $B^{\prime }$ of $V$ as well as $C$ and $C^{\prime }$ of $W$. We are asking now: How can we convert the matrix $M_C^B(f)$ into $M_{C^{\prime }}^{B^{\prime }}(f)$?

In the following, we will consider why the formula in this theorem is correct and how we arrived at it.

From the definition of the matrix of a linear map we know that for all vectors $x\in K^n$, we have $M_C^B(f)x=k_C\circ f\circ k_B^{-1}(x)$ and $M_{C^{\prime }}^{B^{\prime }}(f)x=k_{C^{\prime }}\circ f\circ k_{B^{\prime }}^{-1}(x)$. We can visualize this equation in a diagram:

Representation of the same linear map with respect to different bases as two diagrams

In these two diagrams, it doesn't matter which way you go. For example, it does not matter whether we use $f$ to go directly from $V$ to $W$ or take the detour via $K^n$ and $K^m$. If the same map is constructed along each path, this is referred to as a commutative diagram.

We can join the two diagrams together:

Representation of the same linear map with respect to different bases as a diagram

This diagram is also commutative. That means, if you have a fixed start and end point, it still doesn't matter which path you take in the diagram. If we start at the top left at $K^n$, it doesn't matter which path we use to get to $K^m$ at the bottom left. We can get from $K^n$ to $K^m$ via $x\mapsto M_{C^{\prime }}^{B^{\prime }}(f)x$, or using first $k_B\circ k_{B^{\prime }}^{-1}:K^n\to K^n$, then $x\mapsto M_C^B(f)x$ and finally $k_{C^{\prime }}\circ k_C^{-1}:K^m\to K^m$.

The various compositions are hown in blue and red

Consequently, the map $K^n\to K^m,x\mapsto M_{C^{\prime }}^{B^{\prime }}(f)x$ is equal to the combination of the maps $k_B\circ k_{B^{\prime }}^{-1}$, $x\mapsto M_C^B(f)x$, and $k_{C^{\prime }}\circ k_C^{-1}$. We have now seen that the $x\mapsto M_C^B(f)x$ can be transformed into the map $x\mapsto M_{C^{\prime }}^{B^{\prime }}(f)x$. Originally, however, we wanted to transform the matrix $M_C^B(f)$ into the matrix $M_{C^{\prime }}^{B^{\prime }}(f)$. How do we get from the map $K^n\to K^m,x\mapsto M_{C^{\prime }}^{B^{\prime }}(f)x$ back to the matrix $M_{C^{\prime }}^{B^{\prime }}(f)\in K^{m\times n}$?

The matrix $M_{C^{\prime }}^{B^{\prime }}(f)$ looks complicated. We therefore consider how we can answer this question for a general matrix $A\in K^{m\times n}$. We consider the linear map $L_A:K^n\to K^m,x\mapsto Ax$ associated with $A$. The matrix of $L_A$ with respect to the standard bases of $K^n$ and $K^m$ is again $A$. Let us now plug in the matrix $M_{C^{\prime }}^{B^{\prime }}(f)$ for $A$. The matrix of the linear map $x\mapsto M_{C^{\prime }}^{B^{\prime }}(f)x$ with respect to the standard bases is exactly $M_{C^{\prime }}^{B^{\prime }}(f)$.

As we have already seen, the map $x\mapsto M_{C^{\prime }}^{B^{\prime }}(f)x$ is equal to the combination of the three maps $k_B\circ k_{B^{\prime }}^{-1}$, $x\mapsto M_C^B(f)x$, and $k_{C^{\prime }}\circ k_C^{-1}$. Therefore, the matrix of the combination of $k_B\circ k_{B^{\prime }}^{-1}$, $x\mapsto M_C^B(f)x$, and $k_{C^{\prime }}\circ k_C^{-1}$ corresponds to $M_{C^{\prime }}^{B^{\prime }}(f)$ with regard to the standard bases.

However, we can also determine the matrix of the concatenation in another way. In the article on matrix multiplication, we saw that concatenation between linear maps correspond exactly to the multiplication of the respective matrices. Therefore, we write down the matrices of the concatenated linear maps individually and then multiply them.

• As we have already seen for $M_{C^{\prime }}^{B^{\prime }}(f)$, the matrix of $x\mapsto M_C^B(f)x$ with respect to the standard bases of $K^n$ and $K^m$ is again $M_C^B(f)$.
• We have already derived the matrix of $k_{C^{\prime }}\circ k_C^{-1}$ above; it is $M_{C^{\prime }}^C(\mathrm{id})$. This is exactly the basis change matrix $T_{C^{\prime }}^C$.
• Similarly, the matrix of $k_B\circ k_{B^{\prime }}^{-1}$ is given by the basis change matrix $T_B^{B^{\prime }}=M_B^{B^{\prime }}(\mathrm{id})$.

If we multiply these three matrices, we obtain $M_{C^{\prime }}^{B^{\prime }}(f)$: Vorlage:Einrücken So $M_{C^{\prime }}^{B^{\prime }}(f)$ can be calculated from $M_C^B(f)$ by left multiplication with $T_{C^{\prime }}^C$ and right multiplication with $T_B^{B^{\prime }}$.

Example for a basis change

We now know, how we can convert matrices of a linear map with respect to different bases into each other. Let's look at the example above again. We consider the linear map Vorlage:Einrücken as well as the ordered bases $B=(e_1,e_2)$, $C=((1,1{)}^T,(1,0{)}^T)$, and $C^{\prime }=((1,2{)}^T,(1,0{)}^T)$. We have already calculated the matrix $M_C^B(f)$: Vorlage:Einrücken We want to determine $M_{C^{\prime }}^B(f)$ by matrix multiplication, i.e., by $M_{C^{\prime }}^B(f)=T_{C^{\prime }}^C{M}_C^B(f)T_B^B$. We have to determine $T_B^B$ and $T_{C^{\prime }}^C$. Now, $T_B^B=I_2$, since the basis $B$ does not change. Now let us turn to computing the basis change matrix $T_{C^{\prime }}^C$: We know that $T_{C^{\prime }}^C=M_{C^{\prime }}^C(\mathrm{id})$. In order to determine this matrix, we need to express the basis vectors of $C$ in the basis $C^{\prime }$:

Vorlage:Einrücken Hence, Vorlage:Einrücken Therefore Vorlage:Einrücken You may convince yourself that this result agrees with the result above.

Examples

Basis change for a matrix of a linear map

Consider the bases Vorlage:Einrücken of ${ℝ}^2$, as well as the bases Vorlage:Einrücken of ${ℝ}^3$. Let $f:{ℝ}^2\to {ℝ}^3$ be a map with the following matrix with respect to $B$ and $C$: Vorlage:Einrücken

We want to determine the matrix of $f$ with respect to the bases $B^{\prime }$ and $C^{\prime }$. This can be done by matrix multiplication $M_{C^{\prime }}^{B^{\prime }}(f)=T_{C^{\prime }}^C{M}_C^B(f)T_B^{B^{\prime }}$. To do so, we must first calculate the basis change matrices $T_B^{B^{\prime }}$ and $T_{C^{\prime }}^C$. Mathe für Nicht-Freaks: Vorlage:Beispiel Mathe für Nicht-Freaks: Vorlage:Beispiel