Matrix of a linear map – "Math for Non-Geeks"

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

Definition

Definition (Matrix of a linear map)

Let $K$ be a field, $V$ and $W$ two $K$-vector spaces of dimension $n$ and $m$ respectively. Let $B$ be a basis of $V$ with coordinate mapping $k_B:V\to K^n$ and $C$ a basis of $W$ with coordinate mapping $k_C:W\to K^m$. Further, let $f:V\to W$ be a linear mapping. Define $g:K^n\to K^m$ by $g=k_C\circ f\circ k_B^{-1}$. Now the matrix of $f$ with respect to the bases $B$ and $C$ is given by the corresponding matrix of $g$, i.e., the $i$-th column of the $m\times n$ matrix contains the image $g(e_i)$ of the $i$-th standard basis vector under $g$. We write this as $M_C^B(f)$.

Warning

Note that the matrix $M_C^B(f)$ depends on the selected (ordered) bases $B$ and $C$! If you choose other bases, you generally get a different matrix. This also applies if you only change the order of the base vectors. This is why we use ordered bases.

Hint

The matrix of a linear map is also called the representation matrix or assigned 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,

$k_C(f(b_j))=\left(\begin{array}{c}{a}_{1j}\\ a_{2j}\\ \dots \\ a_{mj}\end{array}\right).$

This means that the $ij$-th entry of $M_C^B(f)$ is given by the entry $a_{ij}$ from the basic representation $f(b_j)=\sum _{i=0}^m{a}_{ij}{c}_i$.

Definition (Matrix of a linear map, alternative definition)

Let $K$ be a field and $V$ and $W$ two finite-dimensional $K$-vector spaces. Let $B=\{b_1,\dots ,b_n\}$ be a basis of $V$ and $C=\{c_1,\dots ,c_m\}$ a basis of $W$. Let $f:V\to W$ be a linear mapping. Further, let $a_{ij}\in K$ be such that $f(b_j)=\sum _{i=1}^m{a}_{ij}{c}_i$ for all $j\in \{1,\dots ,n\}$. Then we define the matrix of $f$ with respect to $B$ and $C$ as the matrix $M_C^B(f)=(a_{ij}{)}_{ij}$.

Hint

The columns of $M_C^B(f)$ are therefore the coordinates with respect to $C$ of the images of the basis vectors of $B$. We can also write this down like this: $M_C^B(f)=\left(\begin{array}{cccc}{k}_C(f(b_1))& k_C(f(b_2))& \dots & k_C(f(b_n))\end{array}\right).$ Here, each entry in the row is a column vector.

Example (Computing the matrix of a linear map)

Let $ℝ[x{]}_{\le 2}$ be the vector space of polynomials of degree at most 2 with coefficients from $ℝ$ and $ℝ[x{]}_{\le 1}$ the vector space of polynomials of degree at most 1 with coefficients from $ℝ$. We define the following linear mapping:

$\begin{array}{cc}f:ℝ[x{]}_{\le 2}& \to ℝ[x{]}_{\le 1}\\ p=p_2{x}^2+p_{1}x+p_0& \mapsto (p_2+p_1)(x+1)+p_0\end{array}$

It is easy to check that $f$ is actually a linear mapping. We have the bases $B=(x^2,x,1)$ of $ℝ[x{]}_{\le 2}$ and $C=(x,1)$ of $ℝ[x{]}_{\le 1}$.

We are looking for the matrix $M_C^B(f)$.

To find it, we calculate the images of the basis vectors from $B$ and express the result in the basis $C$:

$\begin{array}{cc}f(x^2)& =f(1\cdot x^2+0\cdot x+0)=(1+0)(x+1)+0=x+1=1\cdot x+1\cdot 1\\ f(x)& =f(0\cdot x^2+1\cdot x+0)=(0+1)(x+1)+0=x+1=1\cdot x+1\cdot 1\\ f(1)& =f(0\cdot x^2+0\cdot x+1)=(0+0)(x+1)+1=1\cdot 1\end{array}$

The coefficients in front of the basis vectors in $C$ are the entries of the matrix we are looking for. Therefore

$M_C^B(f)=\left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\end{array}\right)$

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

$\begin{array}{cc}f(v)& =f\left(\sum _{j=1}^n{\lambda }_j{b}_j\right)\\ & \downarrow \text{linearity of }f\\ & =\sum _{j=1}^n{\lambda }_{j}f(b_j)\\ & \downarrow \text{definition of the matrix }{M}_C^B(f)=(a_{ij})\\ & =\sum _{j=1}^n{\lambda }_j\left(\sum _{i=1}^m{a}_{ij}{c}_i\right)\\ & \downarrow \text{re-arranging the sum }\\ & =\sum _{i=1}^m\left(\sum _{j=1}^n{a}_{ij}{\lambda }_j\right)c_i\\ \end{array}$

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

${\mu }_i=\sum _{j=1}^n{a}_{ij}{\lambda }_j.$

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

$\left(\begin{array}{c}{\mu }_1\\ {\mu }_2\\ ⋮\\ {\mu }_m\end{array}\right)=\left(\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right)\left(\begin{array}{c}{\lambda }_1\\ {\lambda }_2\\ ⋮\\ {\lambda }_n\end{array}\right)=\left(\begin{array}{c}\sum _{j=1}^n{a}_{1j}{\lambda }_j\\ \sum _{j=1}^n{a}_{2j}{\lambda }_j\\ ⋮\\ \sum _{j=1}^n{a}_{mj}{\lambda }_j\end{array}\right)$

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 $M_C^B(f)$.

$k_C(f(v))=M_C^B(f)k_B(v)$

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,

$f(k_B^{-1}(x))=k_C^{-1}(M_C^B(f)x)$

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

Example (Using the matrix of a linear map)

As above, we consider the linear map

$\begin{array}{cc}f:ℝ[x{]}_{\le 2}& \to ℝ[x{]}_{\le 1}\\ p=p_2{x}^2+p_{1}x+p_0& \mapsto (p_2+p_1)(x+1)+p_0\end{array}$

and the bases $B=(x^2,x,1)$ of $ℝ[x{]}_{\le 2}$ or $C=(x,1)$ of $ℝ[x{]}_{\le 1}$.

We have already calculated the matrix of $f$ with respect to these bases:

$M_C^B(f)=\left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\end{array}\right)$

We can now use this matrix to calculate $f(p)$ for a polynomial $p\in ℝ[x{]}_{\le 2}$. We have seen above that

$k_C(f(p))=M_C^B(f)k_B(p)$

To understand this, let's look at a concrete example: We consider the polynomial $p:=5x^2-3x+10$. First, we need to calculate $k_B(p)$, i.e., the coordinates with respect to the basis $B$. The coordinate vector is formed from the prefactors of the linear combination in the basis $B=(x^2,x,1)$. We have

$p:=5x^2-3x+10$

This allows us to find the coordinate vector

$k_B(p)=\left(\begin{array}{c}5\\ -3\\ 10\end{array}\right)$

We can multiply this vector with the matrix $M_C^B(f)$:

$\begin{array}{cc}& M_C^B(f)k_B(p)\\ =& \left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\end{array}\right)\left(\begin{array}{c}5\\ -3\\ 10\end{array}\right)\\ =& \left(\begin{array}{c}2\\ 12\end{array}\right)\end{array}$

This vector $(2,12{)}^T$ is $k_C(f(p))$, i.e., the coordinate vector of $f(p)$ in the basis $C=(x,1)$. In order to obtain $f(p)$ from this, we must write the coordinates in the vector $(2,12{)}^T$ as prefactors in the linear combination of $C$. Thus

$f(p)=k_C^{-1}\left(\left(\begin{array}{c}2\\ 12\end{array}\right)\right)=2\cdot x+12\cdot 1=2x+12$

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.

Theorem (Matrix of a composition of linear maps)

Let $f:V\to W$ and $g:W\to X$ be linear mappings between finite-dimensional vector spaces. Furthermore, let $B=\{v_1,\dots ,v_m\}$ be a basis of $V$, $C=\{w_1,\dots ,w_n\}$ a basis of $W$ and $D=\{x_1,\dots ,x_s\}$ a basis of $X$. Then

$M_D^B(g\circ f)=M_D^C(g)\cdot M_C^B(f).$

Proof (Matrix of a composition of linear maps)

Let $h=g\circ f$ and let $(h_{ij}{)}_{ij}=M_D^B(h)\in K^{s\times m}$. Further, let $M_C^B(f)=(f_{ij}{)}_{ij}\in K^{n\times m}$ and $M_D^C(g)=(g_{ij}{)}_{ij}\in K^{s\times n}$ be the matrices of $f$ and $g$ respectively.

We now know that the $h_{ij}$ are the unique scalars, satisfying

$h(v_j)=\sum _{i=1}^s{h}_{ij}{x}_i$

for all $j\in \{1,\dots ,m\}$. In order to prove $(h_{ij})=(g_{ij})(f_{ij})$, we need to verify

$h_{ij}=\sum _{k=1}^n{g}_{ik}{f}_{kj}$

Indeed,

$\begin{array}{cc}h(v_j)& =g(f(v_j))\\ & \downarrow \text{definition of }{M}_C^B(f)\\ & =g\left(\sum _{k=1}^n{f}_{kj}{w}_k\right)\\ & \downarrow \text{linearity of }g\\ & =\sum _{k=1}^n{f}_{kj}g(w_k)\\ & \downarrow \text{definition of }{M}_D^C(g)\\ & =\sum _{k=1}^n{f}_{kj}\sum _{i=1}^s{g}_{ik}{x}_i\\ & =\sum _{i=1}^s\left(\sum _{k=1}^n{g}_{ik}{f}_{kj}\right)x_i.\end{array}$

From the uniqueness of the coordinates in the linear combination of $x_i$, we conclude $h_{ij}=\sum _{k=1}^n{g}_{ik}{f}_{kj}$.

Warning

For the reduction rule, it is important that the same ordered basis $C$ of $W$ is chosen in both cases for the matrices representing $f$ and $g$. If $M_D^{\tilde{C}}(g)$ is formed for a different basis $\tilde{C}\ne C$ of $W$, then the reduction rule no longer applies: The equation

$M_D^B(g\circ f)=M_D^{\tilde{C}}(g)\cdot M_C^B(f)$

is generally false. Because representing matrices depend on the order of the basis vectors, this also applies if $\tilde{C}$ is only a rearrangement of $C$.

One-to-one correspondence between matrices and linear maps

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 $M_C^B(f)$:

$\begin{array}{cc}\mathrm{Hom}(V,W)& \to K^{m\times n}\\ f& \mapsto M_C^B(f)\end{array}$

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:

$\begin{array}{cc}\mathrm{Hom}(V,W)& \to \mathrm{Hom}(K^n,K^m)\\ f& \mapsto f^{\prime }:=k_C\circ f\circ k_B^{-1}\end{array}$

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

$\begin{array}{cc}\mathrm{Hom}(K^n,K^m)& \to K^{m\times n}\\ f^{\prime }& \mapsto M(f^{\prime })\end{array}$

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:

$f(b_j)=\sum _{i=1}^m{a}_{ij}{c}_i$

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., $f(b_j)$.

Example (One-to-one correspondence)

We want to better understand the one-to-one correspondence between matrices and linear mappings using an example. The bijection is given by

$\begin{array}{cc}\mathrm{Hom}(V,W)& \to K^{m\times n}\\ f& \mapsto M_C^B(f),\end{array}$

where $B$ is an (ordered) basis of $V$ and $C$ is an (ordered) basis of $W$. We consider the two $ℝ$ vector spaces $V=ℝ[x{]}_{\le 2}$ and $W=ℝ[x{]}_{\le 1}$, i.e., the vector spaces of the polynomials with coefficients from $ℝ$ and degree at most 2 or 1. For the one-to-one correspondence, we still need an ordered basis of $V$ and of $W$. We choose the canonical bases $B:=(x^2,x,1)$ and $C:=(x,1)$. What are the variables $m$ and $n$ in this example? The number $m$ is the dimension of the vector space $W$ and $n$ is the dimension of $V$. So $m=2$ and $n=3$.

We therefore have the bijection

$\begin{array}{cc}\mathrm{Hom}(ℝ[x{]}_{\le 2},ℝ[x{]}_{\le 1})& \to {ℝ}^{2\times 3}\\ f& \mapsto M_C^B(f)\end{array}$

This means every linear map $f$ from $ℝ[x{]}_{\le 2}$ to $ℝ[x{]}_{\le 1}$ provides a $(2\times 3)$ matrix

$M_C^B(f)=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right)$

with coefficients $a_{ij}\in ℝ$. For example, we have seen above that for the linear map

$\begin{array}{cc}f:ℝ[x{]}_{\le 2}& \to ℝ[x{]}_{\le 1}\\ p=p_2{x}^2+p_{1}x+p_0& \mapsto (p_2+p_1)(x+1)+p_0\end{array}$

and the bases $B$ and $C$, we get the corresponding matrix

$M_C^B(f)=\left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\end{array}\right).$

However, the one-to-one correspondence says even more: For every $(2\times 3)$-matrix $A$ with coefficients in $ℝ$ there is a unique linear mapping $f$ from $ℝ[x{]}_{\le 2}$ to $ℝ[x{]}_{\le 1}$, so that $A$ is the mapping matrix of $f$, i.e., $M_C^B(f)=A$.

Hint

If we choose a suitable vector space structure on the set of matrices $K^{m\times n}$, the bijection explained above is even an isomorphism. The vector space structure we have to fix for the matrices is componentwise addition and scalar multiplication. We look at this in more detail in the article "Vector space structure on matrices".

Examples

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

Example (Concrete example)

We consider the linear map

$f:{ℝ}^3\to {ℝ}^2,f\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}2x-3y\\ x-2y+z\end{array}\right)$

The canonical standard basis is selected both in the original space ${ℝ}^3$ and in the target space ${ℝ}^2$:

$B=(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)),C=(\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right))$

We have:

$f\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right),f\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}-3\\ -2\end{array}\right),f\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)$

This means that the matrix of $L$ with respect to the selected bases $B$ and $C$ is:

$M_C^B(f)=\left(\begin{array}{ccc}2& -3& 0\\ 1& -2& 1\end{array}\right)$

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

Example (Concrete example with a different basis)

Again, we consider the linear map $f$ of the above example, i.e.,

$f:{ℝ}^3\to {ℝ}^2,f\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}2x-3y\\ x-2y+z\end{array}\right)$

This time we use the ordered basis in the target space ${ℝ}^2$

$C^{\prime }=(\left(\begin{array}{c}2\\ 1\end{array}\right),\left(\begin{array}{c}1\\ 1\end{array}\right))$

Now,

$f\left(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)=1\cdot \left(\begin{array}{c}2\\ 1\end{array}\right)+0\cdot \left(\begin{array}{c}1\\ 1\end{array}\right),$

$f\left(\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\right)=\left(\begin{array}{c}-3\\ -2\end{array}\right)=-1\cdot \left(\begin{array}{c}2\\ 1\end{array}\right)-1\cdot \left(\begin{array}{c}1\\ 1\end{array}\right),$

$f\left(\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)=-1\cdot \left(\begin{array}{c}2\\ 1\end{array}\right)+2\cdot \left(\begin{array}{c}1\\ 1\end{array}\right)$

This gives the following matrix of $f$ with respect to the bases $B$ and $C^{\prime }$:

$M_{C^{\prime }}^B(f)=\left(\begin{array}{ccc}1& -1& -1\\ 0& -1& 2\end{array}\right)$

We can see that this matrix is not equal to $M_C^B(f)$ from the first example.

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.

Example (Concrete example with a re-arranged basis)

Again, we consider the linear map $f$ of the above example, i.e.,

$f:{ℝ}^3\to {ℝ}^2,f\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}2x-3y\\ x-2y+z\end{array}\right)$

This time we use the reordered standard basis in the target space ${ℝ}^2$

$C^{″}=(\left(\begin{array}{c}0\\ 1\end{array}\right),\left(\begin{array}{c}1\\ 0\end{array}\right))$

Now,

$f\left(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)=1\cdot \left(\begin{array}{c}0\\ 1\end{array}\right)+2\cdot \left(\begin{array}{c}1\\ 0\end{array}\right),$

$f\left(\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\right)=\left(\begin{array}{c}-3\\ -2\end{array}\right)=-2\cdot \left(\begin{array}{c}0\\ 1\end{array}\right)-3\cdot \left(\begin{array}{c}1\\ 0\end{array}\right),$

$f\left(\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)=1\cdot \left(\begin{array}{c}0\\ 1\end{array}\right)+0\cdot \left(\begin{array}{c}1\\ 0\end{array}\right)$

This gives the following matrix of $f$ with respect to the bases $B$ and $C^{″}$:

$M_{C^{″}}^B(f)=\left(\begin{array}{ccc}1& -2& 1\\ 2& -3& 0\end{array}\right)$

We can see that this matrix is neither equal to $M_C^B(f)$ from the first nor to the one from the second example. (In fact, it is the $M_C^B(f)$ from the first example with the rows being swapped.)

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

Example (Concrete example with a different linear map but the same matrix)

Consider the linear map

$F:{ℝ}^3\to {ℝ}^2,F\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}x-y-z\\ -y+2z\end{array}\right)$

We select the standard basis for both the original and target space:

$B=(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)),C=(\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right))$

and, as in the previous examples, we calculate the matrix representing $F$ with respect to these bases as

$M_C^B(F)=\left(\begin{array}{ccc}1& -1& -1\\ 0& -1& 2\end{array}\right).$

This is the same matrix as the $M_{C^{\prime }}^B(f)$ from the previous example. But the linear maps $F$ and $f$ are not identical, because

$f\left(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)\ne \left(\begin{array}{c}1\\ 0\end{array}\right)=F\left(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\right).$

Let us now look at a somewhat more abstract example:

Example (Polynomials of different degrees)

Let $K=ℝ$ and let $V$ be the vector space of polynomials of degree at most 3 with coefficients from $ℝ$. Further, let $W$ the vector space of polynomials of degree at most 2 with coefficients from $ℝ$. We define $f:V\to W$ as the derivative of a polynomial, i.e., for all $p(x)\in V$ we set $p=p_0+p_{1}x+p_2{x}^2+p_3{x}^3\mapsto p^{\prime }=p_1+2p_{2}x+3p_3{x}^2$. When considering the bases: $B=(1,x,x^2,x^3)$ and $C=(1,x,x^2)$, then the following applies:

$f(b_1)=f(1)=0=0\cdot c_1+0\cdot c_2+0\cdot c_3$

$f(b_2)=f(x)=1=1\cdot c_1+0\cdot c_2+0\cdot c_3$

$f(b_3)=f(x^2)=2x=0\cdot c_1+2\cdot c_2+0\cdot c_3$

$f(b_4)=f(x^3)=3x^2=0\cdot c_1+0\cdot c_2+3\cdot c_3$

This gives the following matrix of $f$ with respect to the bases $B$ and $C$:

$M_C^B(f)=\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 3\end{array}\right)$