Basis change via matrices – Serlo

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 $B=\{b_{1},\ldots ,b_{n}\}$ of $V$. Every vector $v\in V$ can therefore be written uniquely as a linear combination of the basis vectors $b_{1},\ldots ,b_{n}$, i.e. $v=\sum _{i=1}^{n}\lambda _{i}b_{i}$ with unique $\lambda _{1},\ldots ,\lambda _{n}\in K$.

We also know that vector spaces usually have more than one basis. Let $C=\{c_{1},\ldots ,c_{n}\}$ 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 $\mu _{1},\ldots ,\mu _{n}\in K$.

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 $B=(e_{1},\ldots ,e_{n})$ as the (ordered) standard basis. Let further $C=(c_{1},\ldots ,c_{n})$ 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 $v=(x_{1},\ldots ,x_{n})^{T}=\sum _{i=1}^{n}x_{i}e_{i}$ 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 $\lambda _{1},\ldots ,\lambda _{n}\in K$. 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 $M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})$ with respect to the standard basis $B=(e_{1},\ldots ,e_{n})$. By using matrix-vector multiplication in $K^{n}$, we then obtain the coordinate vector $(\lambda _{1},\ldots ,\lambda _{n})^{T}$ by multiplying $v=(x_{1},\ldots ,x_{n})^{T}$ from the left by $M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})$.

To calculate the matrix $M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})$, we need to determine $k_{C}(e_{1}),\ldots ,k_{C}(e_{n})$. These will then be the columns of $M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})$. We are therefore looking for the coordinates of $e_{1},\ldots ,e_{n}$ with respect to $C$, so we have to write these as a linear combination of vectors in $C$. This gives us $n$ equations

${\begin{aligned}e_{1}=&\sum _{i=1}^{n}a_{i1}c_{i}\\&\vdots \\e_{n}=&\sum _{i=1}^{n}a_{in}c_{i}\end{aligned}}$

where $a_{ij}$ are the coordinates we are looking for. The coefficients $a_{ij}$ can be determined by solving a linear system of equations.

Example (Change to standard basis)

We will examine this procedure using a concrete example. To do so, we consider $\mathbb {R} ^{3}$ as a vector space with the ordered standard basis

$B=\left({\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\right)$

We also choose the ordered basis $C=(c_{1},c_{2},c_{3})$ as follows:

${\begin{aligned}c_{1}:={\begin{pmatrix}1\\1\\0\end{pmatrix}},\ c_{2}:={\begin{pmatrix}0\\-1\\1\end{pmatrix}},\ c_{3}:={\begin{pmatrix}2\\0\\1\end{pmatrix}}\end{aligned}}$

Each vector in $\mathbb {R} ^{3}$ can be represented in the basis $B$ and the basis $C$ to obtain the above-mentioned coefficients $x_{1},x_{2},x_{3}$ or $\lambda _{1},\lambda _{2},\lambda _{3}$. For example, for the vector $(6,1,3)^{T}$, the coefficients are $x_{1}=6,\ x_{2}=1,\ x_{3}=3$ and $\lambda _{1}=2,\ \lambda _{2}=1,\ \lambda _{3}=2$, because

${\begin{pmatrix}6\\1\\3\end{pmatrix}}=2\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}+1\cdot {\begin{pmatrix}0\\-1\\1\end{pmatrix}}+2\cdot {\begin{pmatrix}2\\0\\1\end{pmatrix}}$

To make it easier to determine the coefficients $\lambda _{1},\lambda _{2},\lambda _{3}$, we express the standard basis in the basis $C$. This means we want to find the coefficients $a_{ij}\in \mathbb {R}$ with

${\begin{aligned}{\begin{pmatrix}1\\0\\0\end{pmatrix}}=a_{11}\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}+a_{21}\cdot {\begin{pmatrix}0\\-1\\1\end{pmatrix}}+a_{31}\cdot {\begin{pmatrix}2\\0\\1\end{pmatrix}}\\[0.3em]{\begin{pmatrix}0\\1\\0\end{pmatrix}}=a_{12}\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}+a_{22}\cdot {\begin{pmatrix}0\\-1\\1\end{pmatrix}}+a_{32}\cdot {\begin{pmatrix}2\\0\\1\end{pmatrix}}\\[0.3em]{\begin{pmatrix}0\\0\\1\end{pmatrix}}=a_{13}\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}+a_{23}\cdot {\begin{pmatrix}0\\-1\\1\end{pmatrix}}+a_{33}\cdot {\begin{pmatrix}2\\0\\1\end{pmatrix}}\end{aligned}}$

By solving the linear system, we can determine and obtain the coefficients:

${\begin{aligned}{\begin{pmatrix}1\\0\\0\end{pmatrix}}&=-1\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}+(-1)\cdot {\begin{pmatrix}0\\-1\\1\end{pmatrix}}+1\cdot {\begin{pmatrix}2\\0\\1\end{pmatrix}}\\[0.3em]{\begin{pmatrix}0\\1\\0\end{pmatrix}}&=2\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}+1\cdot {\begin{pmatrix}0\\-1\\1\end{pmatrix}}+(-1)\cdot {\begin{pmatrix}2\\0\\1\end{pmatrix}}\\[0.3em]{\begin{pmatrix}0\\0\\1\end{pmatrix}}&=2\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}+2\cdot {\begin{pmatrix}0\\-1\\1\end{pmatrix}}+(-1)\cdot {\begin{pmatrix}2\\0\\1\end{pmatrix}}\end{aligned}}$

Then $k_{C}(e_{j})=(a_{1j},a_{2j},\ldots ,a_{nj})^{T}$ for $j=1,\ldots ,n$. This gives us the matrix

$M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})={\begin{pmatrix}a_{11}&a_{12}&\ldots &a_{1n}\\a_{21}&a_{22}&\ldots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n2}&\ldots &a_{nn}\end{pmatrix}}.$

We obtain $M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})y=k_{C}(y)$ for all $y\in K^{n}$. The required coefficients $\lambda _{1},\dots ,\lambda _{n}$ are therefore obtained by

$M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C}){\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}={\begin{pmatrix}\lambda _{1}\\\vdots \\\lambda _{n}\end{pmatrix}}.$

Example (Change to standard basis 2)

For our example above, we can also specify the matrix $M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})$:

$M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C})={\begin{pmatrix}-1&2&2\\-1&1&2\\1&-1&-1\end{pmatrix}}.$

With this matrix, we can also easily calculate the coefficients $\lambda _{1},\lambda _{2},\lambda _{3}$ of the vector $(6,1,3)^{T}$:

${\begin{pmatrix}\lambda _{1}\\\lambda _{2}\\\lambda _{3}\end{pmatrix}}=M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C}){\begin{pmatrix}6\\1\\3\end{pmatrix}}={\begin{pmatrix}-1&2&2\\-1&1&2\\1&-1&-1\end{pmatrix}}{\begin{pmatrix}6\\1\\3\end{pmatrix}}={\begin{pmatrix}2\\1\\2\end{pmatrix}}.$

This means $\lambda _{1}=2,\ \lambda _{2}=1,\ \lambda _{3}=2$, as we have already calculated above.

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 $x_{1},\ldots ,x_{n}$ 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 $x_{1},\ldots ,x_{n}$ 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 $(\lambda _{1},\dots ,\lambda _{n})^{T}$.

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 $M_{\mathrm {Std} }^{\mathrm {Std} }(k_{C}\circ k_{B}^{-1})$. If we remember the article on matrices of linear maps, however, this matrix is just $M_{C}^{B}(\operatorname {id} _{V})$, because $k_{C}\circ k_{B}^{-1}=k_{C}\circ \operatorname {id} _{V}\circ k_{B}^{-1}$.

It also makes intuitive sense that the matrix executing the basis change from $B$ to $C$ is given exactly by $M_{C}^{B}(\operatorname {id} _{V})$ 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 $M_{C}^{B}(\operatorname {id} _{V})$, then we obtain exactly the coordinate vector of $\operatorname {id} _{V}(v)=v$ with respect to $C$, just by definition of the representing matrix. That is,

$k_{C}(v)=M_{C}^{B}(\operatorname {id} _{V})\cdot k_{B}(v)$

for all $v\in V$. The matrix $M_{C}^{B}(\operatorname {id} _{V})$ therefore converts coordinates with respect to $B$ into coordinates with respect to $C$. This is exactly what a basis change matrix does.

Definition

Definition (Basis change matrix)

Let $V$ be a finite-dimensional vector space, and let $B$ and $C$ be two ordered bases of $V$. Then the basis change matrix from $B$ to $C$ is the matrix of the identity map $\operatorname {id_{V}}$ with respect to the bases $B$ and $C$, i.e. $M_{C}^{B}(\operatorname {id_{V}} )$. We call this matrix $T_{C}^{B}$.

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.

Warning

In the literature, the names transformation or transition matrix sometimes also refer to matrices that are not basis change matrices.

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 $f\colon V\to W$ 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 $B'$ or $C'$, we will very likely get a different matrix. We can see this in the following example:

Example (Different matrices of one linear map)

Let us consider the map

$f:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x+y\\2y\end{pmatrix}}$

Let $B=(e_{1},e_{2})$ be the standard basis of $\mathbb {R} ^{2}$. We also consider the ordered bases $C=((1,1)^{T},(1,0)^{T})$ and $C'=((1,2)^{T},(1,0)^{T})$. Then

${\begin{aligned}f(e_{1})&={\begin{pmatrix}1+0\\2\cdot 0\end{pmatrix}}={\begin{pmatrix}1\\0\end{pmatrix}}\\[0.5em]f(e_{2})&={\begin{pmatrix}0+1\\2\cdot 1\end{pmatrix}}={\begin{pmatrix}1\\2\end{pmatrix}}.\end{aligned}}$

Since

${\begin{aligned}{\begin{pmatrix}1\\0\end{pmatrix}}&=0\cdot {\begin{pmatrix}1\\1\end{pmatrix}}+1\cdot {\begin{pmatrix}1\\0\end{pmatrix}}\ {\text{ and}}\\[0.3em]{\begin{pmatrix}1\\2\end{pmatrix}}&=2\cdot {\begin{pmatrix}1\\1\end{pmatrix}}+(-1)\cdot {\begin{pmatrix}1\\0\end{pmatrix}}\end{aligned}}$

the matrix of $f$ with respect to $B$ and $C$ looks as follows:

$M_{C}^{B}(f)={\begin{pmatrix}0&2\\1&-1\end{pmatrix}}$

If we carry out the same calculation with the bases $B$ and $C'$, we get

${\begin{aligned}{\begin{pmatrix}1\\0\end{pmatrix}}&=0\cdot {\begin{pmatrix}1\\2\end{pmatrix}}+1\cdot {\begin{pmatrix}1\\0\end{pmatrix}}\ {\text{ and}}\\[0.3em]{\begin{pmatrix}1\\2\end{pmatrix}}&=1\cdot {\begin{pmatrix}1\\2\end{pmatrix}}+0\cdot {\begin{pmatrix}1\\0\end{pmatrix}}\end{aligned}}.$

This means that the matrix of $f$ with respect to the bases $B$ and $C'$ is

$M_{C'}^{B}(f)={\begin{pmatrix}0&1\\1&0\end{pmatrix}}.$

Therefore, $M_{C}^{B}(f)\neq M_{C'}^{B}(f)$.

Solution of this problem

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

Theorem (Basis change of matrices for linear maps)

Let $f\colon V\to W$ be a linear map and consider the ordered bases $B$ and $B'$ of $V$ as well as $C$ and $C'$ of $W$. Then

$M_{C'}^{B'}(f)=T_{C'}^{C}\cdot M_{C}^{B}(f)\cdot T_{B}^{B'}.$

The matrix representing $f$ with respect to $B'$ and $C'$ is therefore obtained from the matrix of $f$ with respect to $B$ and $C$ by multiplying from the left and from the right with the corresponding basis change matrices.

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 $M_{C'}^{B'}(f)x=k_{C'}\circ f\circ k_{B'}^{-1}(x)$. 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 $x\mapsto M_{C'}^{B'}(f)x$, or using first $k_{B}\circ k_{B'}^{-1}:K^{n}\to K^{n}$, then $x\mapsto M_{C}^{B}(f)x$ and finally $k_{C'}\circ k_{C}^{-1}:K^{m}\to K^{m}$.

The various compositions are hown in blue and red

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

The matrix $M_{C'}^{B'}(f)$ 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 $M_{C'}^{B'}(f)$ for $A$. The matrix of the linear map $x\mapsto M_{C'}^{B'}(f)x$ with respect to the standard bases is exactly $M_{C'}^{B'}(f)$.

As we have already seen, the map $x\mapsto M_{C'}^{B'}(f)x$ is equal to the combination of the three maps $k_{B}\circ k_{B'}^{-1}$, $x\mapsto M_{C}^{B}(f)x$, and $k_{C'}\circ k_{C}^{-1}$. Therefore, the matrix of the combination of $k_{B}\circ k_{B'}^{-1}$, $x\mapsto M_{C}^{B}(f)x$, and $k_{C'}\circ k_{C}^{-1}$ corresponds to $M_{C'}^{B'}(f)$ 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 $M_{C'}^{B'}(f)$, 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 $k_{C'}\circ k_{C}^{-1}$ above; it is $M_{C'}^{C}(\operatorname {id} )$. This is exactly the basis change matrix $T_{C'}^{C}$.
• Similarly, the matrix of $k_{B}\circ k_{B'}^{-1}$ is given by the basis change matrix $T_{B}^{B'}=M_{B}^{B'}(\operatorname {id} )$.

If we multiply these three matrices, we obtain $M_{C'}^{B'}(f)$:

$M_{C'}^{B'}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B'}$

So $M_{C'}^{B'}(f)$ can be calculated from $M_{C}^{B}(f)$ by left multiplication with $T_{C'}^{C}$ and right multiplication with $T_{B}^{B'}$.

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

$f:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\quad {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x+y\\2y\end{pmatrix}}$

as well as the ordered bases $B=(e_{1},e_{2})$, $C=((1,1)^{T},(1,0)^{T})$, and $C'=((1,2)^{T},(1,0)^{T})$. We have already calculated the matrix $M_{C}^{B}(f)$:

$M_{C}^{B}(f)={\begin{pmatrix}0&2\\1&-1\end{pmatrix}}$

We want to determine $M_{C'}^{B}(f)$ by matrix multiplication, i.e., by $M_{C'}^{B}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B}$. We have to determine $T_{B}^{B}$ and $T_{C'}^{C}$. Now, $T_{B}^{B}=I_{2}$, since the basis $B$ does not change. Now let us turn to computing the basis change matrix $T_{C'}^{C}$: We know that $T_{C'}^{C}=M_{C'}^{C}(\operatorname {id} )$. In order to determine this matrix, we need to express the basis vectors of $C$ in the basis $C'$:

${\begin{aligned}{\begin{pmatrix}1\\1\end{pmatrix}}&={\frac {1}{2}}\cdot {\begin{pmatrix}1\\2\end{pmatrix}}+{\frac {1}{2}}\cdot {\begin{pmatrix}1\\0\end{pmatrix}}\\[0.5em]{\begin{pmatrix}1\\0\end{pmatrix}}&=0\cdot {\begin{pmatrix}1\\2\end{pmatrix}}+1\cdot {\begin{pmatrix}1\\0\end{pmatrix}}.\end{aligned}}$

Hence,

$T_{C'}^{C}={\begin{pmatrix}{\frac {1}{2}}&0\\{\frac {1}{2}}&1\end{pmatrix}}.$

Therefore

$M_{C'}^{B}(f)=T_{C'}^{C}M_{C}^{B}(f)={\begin{pmatrix}{\frac {1}{2}}&0\\{\frac {1}{2}}&1\end{pmatrix}}\cdot {\begin{pmatrix}0&2\\1&-1\end{pmatrix}}={\begin{pmatrix}0&1\\1&0\end{pmatrix}}.$

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

$B'=\left({\begin{pmatrix}2\\1\end{pmatrix}},{\begin{pmatrix}1\\1\end{pmatrix}}\right)\quad {\text{and}}\quad B=\left({\begin{pmatrix}0\\-1\end{pmatrix}},{\begin{pmatrix}2\\3\end{pmatrix}}\right)$

of $\mathbb {R} ^{2}$, as well as the bases

$C=\left({\begin{pmatrix}1\\-2\\3\end{pmatrix}},{\begin{pmatrix}0\\0\\-1\end{pmatrix}},{\begin{pmatrix}2\\1\\1\end{pmatrix}}\right)\quad {\text{and}}\quad C'=\left({\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\0\end{pmatrix}}\right)$

of $\mathbb {R} ^{3}$. Let $f:\mathbb {R} ^{2}\to \mathbb {R} ^{3}$ be a map with the following matrix with respect to $B$ and $C$:

$M_{C}^{B}(f)={\begin{pmatrix}5&-3\\2&4\\0&-2\end{pmatrix}}$

We want to determine the matrix of $f$ with respect to the bases $B'$ and $C'$. This can be done by matrix multiplication $M_{C'}^{B'}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B'}$. To do so, we must first calculate the basis change matrices $T_{B}^{B'}$ and $T_{C'}^{C}$.

Example (Basis change in $\mathbb {R} ^{2}$)

Consider the two bases

$B'=\left({\begin{pmatrix}2\\1\end{pmatrix}},{\begin{pmatrix}1\\1\end{pmatrix}}\right)\quad {\text{and}}\quad B=\left({\begin{pmatrix}0\\-1\end{pmatrix}},{\begin{pmatrix}2\\3\end{pmatrix}}\right)$

in $\mathbb {R} ^{2}$ . In order to determine the transition matrix $T_{B}^{B'}$ from $B'$ to $B$, we proceed as follows:

1. We represent the basis vectors of $B'$ as a linear combination of the vectors of $B$:

${\begin{array}{ccrcr}{\begin{pmatrix}2\\1\end{pmatrix}}&=&2\cdot {\begin{pmatrix}0\\-1\end{pmatrix}}&+&1\cdot {\begin{pmatrix}2\\3\end{pmatrix}}\\[0.5em]{\begin{pmatrix}1\\1\end{pmatrix}}&=&{\frac {1}{2}}\cdot {\begin{pmatrix}0\\-1\end{pmatrix}}&+&{\frac {1}{2}}\cdot \ {\begin{pmatrix}2\\3\end{pmatrix}}\end{array}}$

2. We write the determined coefficients of the linear combinations as column vectors in a matrix. This is exactly the transition matrix we are looking for:

$T_{B}^{B'}={\begin{pmatrix}2&{\frac {1}{2}}\\1&{\frac {1}{2}}\end{pmatrix}}$

Example (Basis change in $\mathbb {R} ^{3}$)

We consider the bases

$C=\left({\begin{pmatrix}1\\-2\\3\end{pmatrix}},{\begin{pmatrix}0\\0\\-1\end{pmatrix}},{\begin{pmatrix}2\\1\\1\end{pmatrix}}\right)\quad {\text{and}}\quad C'=\left({\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\0\end{pmatrix}}\right)$

in $\mathbb {R} ^{3}$. We want to calculate the basis change matrix $T_{C'}^{C}$ from $C$ to $C'$. To do this, we represent the basis vectors of $C$ as a linear combination of the vectors of $C'$:

${\begin{array}{ccrcrcr}{\begin{pmatrix}1\\-2\\3\end{pmatrix}}&=&3\cdot {\begin{pmatrix}1\\1\\1\end{pmatrix}}&-&5\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}&+&3\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\\[0.5em]{\begin{pmatrix}0\\0\\-1\end{pmatrix}}&=&-1\cdot {\begin{pmatrix}1\\1\\1\end{pmatrix}}&+&1\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}&+&0\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\\[0.5em]{\begin{pmatrix}2\\1\\1\end{pmatrix}}&=&1\cdot {\begin{pmatrix}1\\1\\1\end{pmatrix}}&+&0\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}&+&1\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\end{array}}$

As above, we obtain the transition matrix $T_{C'}^{C}$ by writing the coefficients of the linear combinations as columns in a matrix:

$T_{C'}^{C}={\begin{pmatrix}3&-1&1\\-5&1&0\\3&0&1\end{pmatrix}}$

Example (Basis change for a matrix of a linear map)

Consider the bases $B'=((2,1)^{T},(1,1)^{T})$ and $B=((0,-1)^{T},(2,3)^{T})$ of $\mathbb {R} ^{2}$ and the bases $C=((1,-2,3)^{T},(0,0,-1)^{T},(2,1,1)^{T})$ and $C'=((1,1,1)^{T},(1,1,0)^{T},(1,0,0)^{T})$ of $\mathbb {R} ^{3}$. Let $f:\mathbb {R} ^{2}\to \mathbb {R} ^{3}$ be a linear map with the following matrix with respect to $B$ and $C$:

$M_{C}^{B}(f)={\begin{pmatrix}5&-3\\2&4\\0&-2\end{pmatrix}}$

We want to determine the matrix of $f$ with respect to the bases $B'$ and $C'$. We do this via matrix multiplication $M_{C'}^{B'}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B'}$. In the previous examples, we have already determined $T_{B}^{B'}$ and $T_{C'}^{C}$. So we can simply calculate:

${\begin{aligned}&M_{C'}^{B'}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B'}={\begin{pmatrix}3&-1&1\\-5&1&0\\3&0&1\end{pmatrix}}\cdot {\begin{pmatrix}5&-3\\2&4\\0&-2\end{pmatrix}}\cdot {\begin{pmatrix}2&{\frac {1}{2}}\\1&{\frac {1}{2}}\end{pmatrix}}\\[0.5em]=&{\begin{pmatrix}3&-1&1\\-5&1&0\\3&0&1\end{pmatrix}}\cdot {\begin{pmatrix}7&1\\8&3\\-2&-1\end{pmatrix}}={\begin{pmatrix}11&-1\\-27&-2\\19&2\end{pmatrix}}\end{aligned}}$

The matrix of $f$ with respect to the bases $B'$ and $C'$ is therefore

$M_{C'}^{B'}(f)={\begin{pmatrix}11&-1\\-27&-2\\19&2\end{pmatrix}}.$

Exercises

Exercise

We consider the following linear map

$f\colon \mathbb {R} [x]_{\leq 3}\to \mathbb {R} ,\quad p=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\mapsto f(p)=a_{0}+a_{1}+a_{2}+a_{3}$

as well as the bases $B=\{1,x,x^{2},x^{3}\}$ and $B'=\{1,x+1,x^{2}+x,x^{3}+x^{2}\}$ of $\mathbb {R} [x]_{\leq 3}$ and the bases $C=\{1\}$ and $C'=\{2\}$ of $\mathbb {R}$.

1. Calculate the matrix of $f$ with respect to $B$ and $C$, as well as the matrix with respect to $B'$ and $C'$.
2. Calculate the basis change matrix from $B$ to $B'$, and vice versa from $B'$ to $B$.
3. Calculate the basis change matrix from $C$ to $C'$, and vice versa from $C'$ to $C$.
4. Verify that you can calculate the matrix $M_{C'}^{B'}(f)$ from the matrix $M_{C}^{B}(f)$ using the basis change matrices.

Solution

Solution sub-exercise 1:

We calculate the images of the basis vectors:

${\begin{aligned}f(1)=1=1\cdot 1\\f(x)=1=1\cdot 1\\f(x^{2})=1=1\cdot 1\\f(x^{3})=1=1\cdot 1\end{aligned}}$

The corresponding matrix of $f$ is therefore

$M_{C}^{B}(f)={\begin{pmatrix}1&1&1&1\end{pmatrix}}$

As above, we calculate the images of the basis vectors:

${\begin{aligned}f(1)&=1={\frac {1}{2}}\cdot 2\\f(x+1)&=2=1\cdot 2\\f(x^{2}+x)&=2=1\cdot 2\\f(x^{3}+x^{2})&=2=1\cdot 2\end{aligned}}$

In the second step, we express the images in the basis $C'=\{2\}$. The desired matrix is therefore

$M_{C'}^{B'}(f)={\begin{pmatrix}{\frac {1}{2}}&1&1&1\end{pmatrix}}$

Solution sub-exercise 2:

To determine the basis change matrix $T_{B}^{B'}$ from $B'$ to $B$, we first represent the basis vectors of $B'$ as a linear combination of the vectors of $B$:

${\begin{aligned}1&=1\cdot 1+0\cdot x+0\cdot x^{2}+0\cdot x^{3}\\x+1&=1\cdot 1+1\cdot x+0\cdot x^{2}+0\cdot x^{3}\\x^{2}+x&=0\cdot 1+1\cdot x+1\cdot x^{2}+0\cdot x^{3}\\x^{3}+x^{2}&=0\cdot 1+0\cdot x+1\cdot x^{2}+1\cdot x^{3}\end{aligned}}$

The coefficients of the linear combinations are the column vectors of the matrix we are looking for:

$T_{B}^{B'}={\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{pmatrix}}$

Likewise, we can calculate the basis change matrix $T_{B'}^{B}$ from $B'$ to $B$. Alternatively, we can also calculate the inverse matrix of $T_{B}^{B'}$:

$T_{B'}^{B}=(T_{B}^{B'})^{-1}={\begin{pmatrix}1&-1&1&-1\\0&1&-1&1\\0&0&1&-1\\0&0&0&1\end{pmatrix}}$

Solution sub-exercise 3:

As in the previous part, we represent the basis vectors of $C'$ as a linear combination of the vectors of $C$:

$2=2\cdot 1$

Hence

$T_{C}^{C'}={\begin{pmatrix}2\end{pmatrix}}$

For the inverse, we have

$1={\frac {1}{2}}\cdot 2$

Thus

$T_{C'}^{C}={\begin{pmatrix}{\frac {1}{2}}\end{pmatrix}}$

Solution sub-exercise 4:

According to the above formula, we should have $M_{C'}^{B'}(f)=T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B'}$. Indeed, the left-hand side is

$M_{C'}^{B'}(f)={\begin{pmatrix}{\frac {1}{2}}&1&1&1\end{pmatrix}}$

On the right-hand side, we get

${\begin{aligned}&T_{C'}^{C}M_{C}^{B}(f)T_{B}^{B'}\\[0.5em]=&{\begin{pmatrix}{\frac {1}{2}}\end{pmatrix}}{\begin{pmatrix}1&1&1&1\end{pmatrix}}{\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{pmatrix}}\\[0.5em]=&{\begin{pmatrix}{\frac {1}{2}}\end{pmatrix}}{\begin{pmatrix}1&2&2&2\end{pmatrix}}\\[0.5em]=&{\begin{pmatrix}{\frac {1}{2}}&1&1&1\end{pmatrix}}\end{aligned}}$

And in fact, both matrices agree!