Serlo: EN: Matrix of a linear map

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In this article, we will learn how to describe linear maps between arbitrary finite-dimensional vector spaces using matrices. The matrix representing such a linear mapping $f:V\to W$ depends on the choice of bases in $V$ and in $W$. Their columns are the coordinates of the images of the base vectors of $V$.

Generalization to abstract vector spaces

In the article on introduction to matrices, we saw how we can describe a linear mapping $K^n\to K^m$ using a matrix. In this way, we can specify and classify linear mappings between $K^n$ and $K^m$ quite easily. Can we also find such a description for linear mappings between general vector spaces?

Formally speaking, we care asking: Given two finite-dimensional $K$ vector spaces $V$ and $W$, how can we completely describe a linear mapping $f:V\to W$?

To answer this question, we can try to trace it back to the case of $K^n$ and $K^m$. In the article on isomorphisms we have seen that every finite-dimensional vector space is isomorphic to $K^n$. This means $V\cong K^n$ and $W\cong K^m$, where we set $n=\dim (V)$ and $m=\dim (W)$. This isomorphism works as follows: We choose an ordered basis $B=(b_1,\dots ,b_n)$ from $V$. By representing a vector in $V$ with respect to $B$, we obtain the coordinate mapping $k_B:V\to K^n$, which maps $v={\lambda }_1{b}_1+\dots +{\lambda }_n{b}_n\in V$ to $({\lambda }_1,\dots ,{\lambda }_n{)}^T\in K^n$. In the same way, we obtain the isomorphism $k_C:W\to K^m$ after choosing a basis $C$ of $W$. It is important here that $B$ and $C$ are ordered bases, as we would get a different mapping for different arrangements of the basis vectors.

Using these isomorphisms, we can turn our mapping $f:V\to W$ into a mapping $f^{\prime }:K^n\to K^m$: We set $f^{\prime }=k_C\circ f\circ k_B^{-1}$

Shifting a linear map in coordinate space

We can assign a matrix $M$ to this mapping $f^{\prime }$ as described in the article Introduction to matrices.

Have we achieved our goal? If so, we can reconstruct the mapping $f$ from $M$. From the article introduction to matrices, we already know that we can reconstruct the mapping $f^{\prime }:K^n\to K^m$ from $M$ using the induced mapping. Now $k_B$ and $k_C$ are isomorphisms. This means that we can reconstruct $f$ from $f^{\prime }$ via $k_C^{-1}\circ f^{\prime }\circ k_B=k_C^{-1}\circ k_C\circ f\circ k_B^{-1}\circ k_B=f$.

We can therefore call $M$ the matrix assigned to $f$. However, we have to be careful with this name: the matrix depends on the choice of the two ordered bases $B$ of $V$ and $C$ of $W$. This means we have actually found several ways to construct a matrix from $f$. Only after fixing the bases $B$ and $C$ have we found a unique way to get a matrix for $f$. Thus, the matrix $M$ constructed above should actually be called “the matrix assigned to $f$ with respect to the bases $B$ and $C$”. Appropriately, we can denote $M$ by $M_C^B(f)$. By construction, this matrix fills exactly the bottom row in the following diagram:

Diagram characterizing the transformation matrix

Calculating with matrices of linear maps

Computing the matrix of a linear map

Relationship between the elements $b_j,e_j,f(b_j)$ and ${\underset{\_}{a}}_j=(a_{1j},\dots ,a_{mj}{)}^T$.

How can we find the corresponding matrix for $f:V\to W$? That is, how can we specifically calculate the entries of the matrix $M_C^B(f)$?

The $j$-th column vector of the matrix $M_C^B(f)$ is given by $(k_C\circ f\circ k_B^{-1})(e_j)$. We therefore want to determine this vector. Now, $(k_C\circ f\circ k_B^{-1})(e_j)=k_C(f(k_B^{-1}(e_j)))$. The defining property of the coordinate mapping $k_B$ is that it maps the basis vector $b_j$ to $e_j$. Therefore, $k_B^{-1}(e_j)=b_j$. Thus, the $j$-th column of $M_C^B(f)$ is the vector $(k_C(f(b_j))$. To find out how $k_C$ represents the vector $f(b_j)$, we need to represent this vector in the basis $C$. There are scalars $a_{1j},a_{2j},\dots ,a_{mj}\in K$, so that $f(v_j)=\sum _{i=0}^m{a}_{ij}{c}_i$. Then, Vorlage:Einrücken This means that the $ij$-th entry of $M_C^B(f)$ is given by the entry $a_{ij}$ from the basic representation $f(b_j)=\sum _{i=0}^m{a}_{ij}{c}_i$.

Using the matrix of a linear map

Now we know how to calculate the matrix of $f$ with respect to the bases $B=\{b_1,\dots ,b_n\}$ and $C=\{c_1,\dots ,c_m\}$. What can we use this matrix for?

This matrix can be used to calculate the image vector $f(v)$ of each $v\in V$. To do so, we first represent $v$ with respect to the basis $B$ of $V$, i.e., $v={\lambda }_1{b}_1+{\lambda }_2{b}_2+\cdots +{\lambda }_n{b}_n$. We denote the entries of the mapping matrix with $M_C^B(f)=(a_{ij})$. Then we have Vorlage:Einrücken We therefore obtain a representation of the vector $f(v)=\sum _{i=1}^m{\mu }_i{c}_i$ as a linear combination of the basis vectors of $C$, with coordinates Vorlage:Einrücken Using the matrix multiplication with a vector (“row times column”) we can also express this as follows: Vorlage:Einrücken Using the matrix of $f$, we therefore obtain the coordinate vector $k_B(v)=({\lambda }_1,\dots ,{\lambda }_n)$ of $v$ from the coordinate vector $k_C(f(v))$ of $f(v)$: We multiply $k_B(v)$ from the left by the matrix $M_C^B(f)$. Vorlage:Einrücken The equation states that, starting from a vector $v\in V$, the red and blue paths in the diagram for the matrix to be displayed provide the same result.

The defining diagram for a matrix of a linear map

Instead of starting with a vector $v\in V$, we can also start with any vector $x=(x_1,\dots ,x_n{)}^T\in K^n$. Then $x$ is the coordinate vector of $k_B^{-1}(x)=x_1\cdot b_1+\dots +x_n\cdot b_n=:v$. We can also understand the product $y=M_C^B(f)x\in K^m$ as a coordinate vector of $k_C^{-1}(y)=y_1\cdot c_1+\dots +y_m\cdot c_m$. From the diagram, we know that $y$ is the coordinate vector of $f(v)$. Therefore, Vorlage:Einrücken Here we have used the fact that the coordinate mappings are isomorphisms, so we can also reverse the arrows of $k_B$ and $k_C$ in the diagram. The equation states that the red and blue paths in the following diagram give the same result:

The defining diagram of a mapping matrix with inverted coordinate mapping

Matrix of a composition of linear maps

In the following theorem we show that the combination of linear mappings corresponds to the multiplication of their representing matrices.

One-to-one correspondence between matrices and linear mapsVorlage:Anker

We can uniquely assign a matrix $M_C^B(f)$ to a linear map $f$ after a fixed choice of ordered bases $B$ and $C$. This gives us a function that sends a mapping $f$ to its associated matrix $M_C^B(f)$: Vorlage:Einrücken In this formula, $\mathrm{Hom}(V,W)$ is the set of all linear maps from $V$ to $W$ and $K^{m\times n}$ is the set of all $m\times n$ matrices.

How did we arrive at the assignment of the matrix $M_C^B(f)$ to the linear map $f$? We first found a unique mapping $f^{\prime }:K^n\to K^m$ for $f$ using the bases $B$ and $C$ and then determined the matrix assigned to $f^{\prime }$. The mapping $f^{\prime }$ is defined by the coordinate mappings: $f^{\prime }=k_C\circ f\circ k_B^{-1}$. So we have the assignment: Vorlage:Einrücken Because $k_C$ and $k_B$ are bijections, we can also get a unique $f:V\to W$ from an $f^{\prime }:K^n\to K^m$, to which $f^{\prime }$ is assigned. All we have to do is to set $f:=k_C^{-1}\circ f^{\prime }\circ k_B$.

So we have a bijection between $\mathrm{Hom}(V,W)$ and $\mathrm{Hom}(K^n,K^m)$.

The assignment Vorlage:Einrücken is a bijection, as we already saw in the introduction article to matrices.

Therefore, $\mathrm{Hom}(V,W)\to K^{m\times n}$ is also a bijection, because it is the combination of the two bijections $\mathrm{Hom}(V,W)\to \mathrm{Hom}(K^n,K^m)$ and $\mathrm{Hom}(V,W)\to K^{m\times n}$. But what does the inverse of the bijection $\mathrm{Hom}(V,W)\to K^{m\times n}$ look like?

The inverse mapping $K^{m\times n}\to \mathrm{Hom}(V,W)$ sends a matrix $A\in K^{m\times n}$ to a linear map $f:V\to W$ such that $M_C^B(f)=A$. Let $B=(b_1,\dots ,b_n)$ and $C=(c_1,\dots ,c_m)$ be ordered bases of $V$ and $W$ and $A=(a_{ij})$, i.e., $a_{ij}$ is the $i,j$-th component of the matrix $A$. Because $M_C^B(f)=A$, the following must hold: Vorlage:Einrücken Because of the principle of linear continuation, $f$ is already completely defined. Here, we see that $a_{ij}$ is the weight of $c_i$ in $f(b_j)$. Intuitively, the $j$-th column of the mapping matrix again stores the image of the $j$-th basis vector, i.e., $f(b_j)$.

Examples

We calculate the matrix representing a specific linear map ${ℝ}^3\to {ℝ}^2$ with respect to the standard basis.

Now let's look at the same linear map, but a different basis in the target space.

From the two previous examples, we see that the matrix representing a linear map depends on the chosen bases. It is important that we consider ordered bases: The representing matrix also depends on the order of the basis vectors.

Conversely, different mappings can also have the same mapping matrix if they are evaluated for different bases:

Let us now look at a somewhat more abstract example:

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