Matrix of a linear map

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 ${\displaystyle f\colon V\to W}$ depends on the choice of bases in ${\displaystyle V}$ and in ${\displaystyle W}$. Their columns are the coordinates of the images of the base vectors of ${\displaystyle V}$.

In the article on introduction to matrices, we saw how we can describe a linear mapping ${\displaystyle K^{n}\to K^{m}}$ using a matrix. In this way, we can specify and classify linear mappings between ${\displaystyle K^{n}}$ and ${\displaystyle K^{m}}$ 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 ${\displaystyle K}$ vector spaces ${\displaystyle V}$ and ${\displaystyle W}$, how can we completely describe a linear mapping ${\displaystyle f\colon V\to W}$?

To answer this question, we can try to trace it back to the case of ${\displaystyle K^{n}}$ and ${\displaystyle K^{m}}$. In the article on isomorphisms we have seen that every finite-dimensional vector space is isomorphic to ${\displaystyle K^{n}}$. This means ${\displaystyle V\cong K^{n}}$ and ${\displaystyle W\cong K^{m}}$, where we set ${\displaystyle n=\dim(V)}$ and ${\displaystyle m=\dim(W)}$. This isomorphism works as follows: We choose an ordered basis ${\displaystyle B=(b_{1},\dots ,b_{n})}$ from ${\displaystyle V}$. By representing a vector in ${\displaystyle V}$ with respect to ${\displaystyle B}$, we obtain the coordinate mapping ${\displaystyle k_{B}\colon V\to K^{n}}$, which maps ${\displaystyle v=\lambda _{1}b_{1}+\dots +\lambda _{n}b_{n}\in V}$ to ${\displaystyle (\lambda _{1},\dots ,\lambda _{n})^{T}\in K^{n}}$. In the same way, we obtain the isomorphism ${\displaystyle k_{C}\colon W\to K^{m}}$ after choosing a basis ${\displaystyle C}$ of ${\displaystyle W}$. It is important here that ${\displaystyle B}$ and ${\displaystyle C}$ 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 ${\displaystyle f\colon V\to W}$ into a mapping ${\displaystyle f'\colon K^{n}\to K^{m}}$: We set ${\displaystyle f'=k_{C}\circ f\circ k_{B}^{-1}}$

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

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

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

How can we find the corresponding matrix for ${\displaystyle f\colon V\to W}$? That is, how can we specifically calculate the entries of the matrix ${\displaystyle M_{C}^{B}(f)}$?

The ${\displaystyle j}$-th column vector of the matrix ${\displaystyle M_{C}^{B}(f)}$ is given by ${\displaystyle (k_{C}\circ f\circ k_{B}^{-1})(e_{j})}$. We therefore want to determine this vector. Now, ${\displaystyle (k_{C}\circ f\circ k_{B}^{-1})(e_{j})=k_{C}(f(k_{B}^{-1}(e_{j})))}$. The defining property of the coordinate mapping ${\displaystyle k_{B}}$ is that it maps the basis vector ${\displaystyle b_{j}}$ to ${\displaystyle e_{j}}$. Therefore, ${\displaystyle k_{B}^{-1}(e_{j})=b_{j}}$. Thus, the ${\displaystyle j}$-th column of ${\displaystyle M_{C}^{B}(f)}$ is the vector ${\displaystyle (k_{C}(f(b_{j}))}$. To find out how ${\displaystyle k_{C}}$ represents the vector ${\displaystyle f(b_{j})}$, we need to represent this vector in the basis ${\displaystyle C}$. There are scalars ${\displaystyle a_{1j},a_{2j},\ldots ,a_{mj}\in K}$, so that ${\displaystyle f(v_{j})=\sum _{i=0}^{m}a_{ij}c_{i}}$. Then,

This means that the ${\displaystyle ij}$-th entry of ${\displaystyle M_{C}^{B}(f)}$ is given by the entry ${\displaystyle a_{ij}}$ from the basic representation ${\displaystyle f(b_{j})=\sum _{i=0}^{m}a_{ij}c_{i}}$.

Now we know how to calculate the matrix of ${\displaystyle f}$ with respect to the bases ${\displaystyle B=\{b_{1},\ldots ,b_{n}\}}$ and ${\displaystyle C=\{c_{1},\ldots ,c_{m}\}}$. What can we use this matrix for?

This matrix can be used to calculate the image vector ${\displaystyle f(v)}$ of each ${\displaystyle v\in V}$. To do so, we first represent ${\displaystyle v}$ with respect to the basis ${\displaystyle B}$ of ${\displaystyle V}$, i.e., ${\displaystyle v=\lambda _{1}b_{1}+\lambda _{2}b_{2}+\cdots +\lambda _{n}b_{n}}$. We denote the entries of the mapping matrix with ${\displaystyle M_{C}^{B}(f)=(a_{ij})}$. Then we have

We therefore obtain a representation of the vector ${\displaystyle f(v)=\sum _{i=1}^{m}\mu _{i}c_{i}}$ as a linear combination of the basis vectors of ${\displaystyle C}$, with coordinates

Using the matrix multiplication with a vector (“row times column”) we can also express this as follows:

Using the matrix of ${\displaystyle f}$, we therefore obtain the coordinate vector ${\displaystyle k_{B}(v)=(\lambda _{1},\ldots ,\lambda _{n})}$ of ${\displaystyle v}$ from the coordinate vector ${\displaystyle k_{C}(f(v))}$ of ${\displaystyle f(v)}$: We multiply ${\displaystyle k_{B}(v)}$ from the left by the matrix ${\displaystyle M_{C}^{B}(f)}$.

The equation states that, starting from a vector ${\displaystyle v\in V}$, the red and blue paths in the diagram for the matrix to be displayed provide the same result.

Instead of starting with a vector ${\displaystyle v\in V}$, we can also start with any vector ${\displaystyle x=(x_{1},\ldots ,x_{n})^{T}\in K^{n}}$. Then ${\displaystyle x}$ is the coordinate vector of ${\displaystyle k_{B}^{-1}(x)=x_{1}\cdot b_{1}+\ldots +x_{n}\cdot b_{n}=:v}$. We can also understand the product ${\displaystyle y=M_{C}^{B}(f)x\in K^{m}}$ as a coordinate vector of ${\displaystyle k_{C}^{-1}(y)=y_{1}\cdot c_{1}+\ldots +y_{m}\cdot c_{m}}$. From the diagram, we know that ${\displaystyle y}$ is the coordinate vector of ${\displaystyle f(v)}$. Therefore,

Here we have used the fact that the coordinate mappings are isomorphisms, so we can also reverse the arrows of ${\displaystyle k_{B}}$ and ${\displaystyle k_{C}}$ in the diagram. The equation states that the red and blue paths in the following diagram give the same result:

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

We can uniquely assign a matrix ${\displaystyle M_{C}^{B}(f)}$ to a linear map ${\displaystyle f}$ after a fixed choice of ordered bases ${\displaystyle B}$ and ${\displaystyle C}$. This gives us a function that sends a mapping ${\displaystyle f}$ to its associated matrix ${\displaystyle M_{C}^{B}(f)}$:

In this formula, ${\displaystyle \operatorname {Hom} (V,W)}$ is the set of all linear maps from ${\displaystyle V}$ to ${\displaystyle W}$ and ${\displaystyle K^{m\times n}}$ is the set of all ${\displaystyle m\times n}$ matrices.

How did we arrive at the assignment of the matrix ${\displaystyle M_{C}^{B}(f)}$ to the linear map ${\displaystyle f}$? We first found a unique mapping ${\displaystyle f'\colon K^{n}\to K^{m}}$ for ${\displaystyle f}$ using the bases ${\displaystyle B}$ and ${\displaystyle C}$ and then determined the matrix assigned to ${\displaystyle f'}$. The mapping ${\displaystyle f'}$ is defined by the coordinate mappings: ${\displaystyle f'=k_{C}\circ f\circ k_{B}^{-1}}$. So we have the assignment:

Because ${\displaystyle k_{C}}$ and ${\displaystyle k_{B}}$ are bijections, we can also get a unique ${\displaystyle f\colon V\to W}$ from an ${\displaystyle f'\colon K^{n}\to K^{m}}$, to which ${\displaystyle f'}$ is assigned. All we have to do is to set ${\displaystyle f:=k_{C}^{-1}\circ f'\circ k_{B}}$.

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

The assignment

is a bijection, as we already saw in the introduction article to matrices.

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

The inverse mapping ${\displaystyle K^{m\times n}\to \operatorname {Hom} (V,W)}$ sends a matrix ${\displaystyle A\in K^{m\times n}}$ to a linear map ${\displaystyle f\colon V\to W}$ such that ${\displaystyle M_{C}^{B}(f)=A}$. Let ${\displaystyle B=(b_{1},\dots ,b_{n})}$ and ${\displaystyle C=(c_{1},\dots ,c_{m})}$ be ordered bases of ${\displaystyle V}$ and ${\displaystyle W}$ and ${\displaystyle A=(a_{ij})}$, i.e., ${\displaystyle a_{ij}}$ is the ${\displaystyle i,j}$-th component of the matrix ${\displaystyle A}$. Because ${\displaystyle M_{C}^{B}(f)=A}$, the following must hold:

Because of the principle of linear continuation, ${\displaystyle f}$ is already completely defined. Here, we see that ${\displaystyle a_{ij}}$ is the weight of ${\displaystyle c_{i}}$ in ${\displaystyle f(b_{j})}$. Intuitively, the ${\displaystyle j}$-th column of the mapping matrix again stores the image of the ${\displaystyle j}$-th basis vector, i.e., ${\displaystyle f(b_{j})}$.

We calculate the matrix representing a specific linear map ${\displaystyle \mathbb {R} ^{3}\to \mathbb {R} ^{2}}$ 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: