6.6: The matrix of a linear map

Now we will see that every linear map $T \in \mathcal{L}(V, W)$, with $V$ and $W$ finite-dimensional vector spaces, can be encoded by a matrix, and, vice versa, every matrix defines such a linear map.

Let $V$ and $W$ be finite-dimensional vector spaces, and let $T:V\to W$ be a linear map. Suppose that $(v_1,\ldots,v_n)$ is a basis of $V$ and that $(w_1,\ldots,w_m)$ is a basis for $W$. We have seen in Theorem 6.1.3 that $T$ is uniquely determined by specifying the vectors $Tv_1,\ldots, Tv_n\in W$. Since $(w_1,\ldots,w_m)$ is a basis of $W$, there exist unique scalars $a_{ij}\in\mathbb{F}$ such that
$$\begin{equation}\label{eq:Tv} Tv_j = a_{1j} w_1 + \cdots + a_{mj} w_m \quad \text{for \(1\le j\le n \).} \tag{6.6.1} \end{equation}$$
We can arrange these scalars in an $m\times n$ matrix as follows:
$$\begin{equation*} M(T) = \begin{bmatrix} a_{11} & \ldots & a_{1n}\\ \vdots && \vdots\\ a_{m1} & \ldots & a_{mn} \end{bmatrix}. \end{equation*}$$
Often, this is also written as $A=(a_{ij})_{1\le i\le m,1\le j\le n}$. As in Section A.1.1, the set of all $m\times n$ matrices with entries in $\mathbb{F}$ is denoted by $\mathbb{F}^{m\times n}$.

Remark 6.6.1. It is important to remember that $M(T)$ not only depends on the linear map $T$ but also on the choice of the basis $(v_1,\ldots,v_n)$ for $V$ and the choice of basis $(w_1,\ldots,w_m)$ for $W$. The $j^{\text{th}}$ column of $M(T)$ contains the coefficients of the $j^{\text{th}}$ basis vector $v_j$ when expanded in terms of the basis $(w_1,\ldots,w_m)$, as in Equation 6.6.1.

Example 6.6.2. Let $T:\mathbb{R}^2\to \mathbb{R}^2$ be the linear map given by $T(x,y)=(ax+by,cx+dy)$ for some $a,b,c,d\in\mathbb{R}$. Then, with respect to the canonical basis of $\mathbb{R}^2$ given by $((1,0),(0,1))$, the corresponding matrix is
$$\begin{equation*} M(T) = \begin{bmatrix} a&b\\ c&d \end{bmatrix} \end{equation*}$$
since $T(1,0) = (a,c)$ gives the first column and $T(0,1)=(b,d)$ gives the second column.

More generally, suppose that $V=\mathbb{F}^n$ and $W=\mathbb{F}^m$, and denote the standard basis for $V$ by $(e_1,\ldots,e_n)$ and the standard basis for $W$ by $(f_1,\ldots,f_m)$. Here, $e_i$ (resp. $f_i$) is the $n$-tuple (resp. $m$-tuple) with a one in position $i$ and zeroes everywhere else. Then the matrix $M(T)=(a_{ij})$ is given by

$$\begin{equation*} a_{ij} = (Te_j)_i, \end{equation*}$$
where $(Te_j)_i$ denotes the $i^{\text{th}}$ component of the vector $Te_j$.

Example 6.6.3. Let $T:\mathbb{R}^2\to\mathbb{R}^3$ be the linear map defined by $T(x,y)=(y,x+2y,x+y)$. Then, with respect to the standard basis, we have $T(1,0)=(0,1,1)$ and $T(0,1)=(1,2,1)$ so that
$$\begin{equation*} M(T) = \begin{bmatrix} 0&1\\ 1& 2 \\ 1&1 \end{bmatrix}. \end{equation*}$$
However, if alternatively we take the bases $((1,2),(0,1))$ for $\mathbb{R}^2$ and
$((1,0,0),(0,1,0),(0,0,1))$ for $\mathbb{R}^3$, then $T(1,2)=(2,5,3)$ and $T(0,1)=(1,2,1)$ so that
$$\begin{equation*} M(T) = \begin{bmatrix} 2&1\\ 5&2 \\ 3&1 \end{bmatrix}. \end{equation*}$$

Example 6.6.4. Let $S:\mathbb{R}^2\to \mathbb{R}^2$ be the linear map $S(x,y)=(y,x)$. With respect to the basis $((1,2),(0,1))$ for $\mathbb{R}^2$, we have
$$\begin{equation*} S(1,2) = (2,1) = 2(1,2) -3(0,1) \quad \text{and} \quad S(0,1) = (1,0) = 1(1,2)-2(0,1), \end{equation*}$$
and so
$$M(S) = \begin{bmatrix} 2&1\\- 3& -2 \end{bmatrix}.$$

Given vector spaces $V$ and $W$ of dimensions $n$ and $m$, respectively, and given a fixed choice of bases, note that there is a one-to-one correspondence between linear maps in $\mathcal{L}(V,W)$ and matrices in $\mathbb{F}^{m\times n}$. If we start with the linear map $T$, then the matrix $M(T)=A=(a_{ij})$ is defined via Equation 6.6.1. Conversely, given the matrix $A=(a_{ij})\in \mathbb{F}^{m\times n}$, we can define a linear map $T:V\to W$ by setting

$$Tv_j = \sum_{i=1}^m a_{ij} w_i.$$

Recall that the set of linear maps $\mathcal{L}(V,W)$ is a vector space. Since we have a one-to-one correspondence between linear maps and matrices, we can also make the set of matrices $\mathbb{F}^{m\times n}$ into a vector space. Given two matrices $A=(a_{ij})$ and $B=(b_{ij})$ in $\mathbb{F}^{m\times n}$ and given a scalar $\alpha\in \mathbb{F}$, we define the matrix addition and scalar multiplication component-wise:

$$\begin{equation*} \begin{split} A+B &= (a_{ij}+b_{ij}),\\ \alpha A &= (\alpha a_{ij}). \end{split} \end{equation*}$$

Next, we show that the composition of linear maps imposes a product on matrices, also called matrix multiplication. Suppose $U,V,W$ are vector spaces over $\mathbb{F}$ with bases $(u_1,\ldots,u_p)$, $(v_1,\ldots,v_n)$ and $(w_1,\ldots,w_m)$, respectively. Let $S:U\to V$ and $T:V\to W$ be linear maps. Then the product is a linear map $T\circ S:U\to W$.

Each linear map has its corresponding matrix $M(T)=A, M(S)=B$ and $M(TS)=C$. The question is whether $C$ is determined by $A$ and $B$. We have, for each $j\in \{1,2,\ldots p\}$, that

$$\begin{equation*} \begin{split} (T\circ S) u_j &= T(b_{1j}v_1 + \cdots + b_{nj} v_n) = b_{1j} Tv_1 + \cdots + b_{nj} Tv_n\\ &= \sum_{k=1}^n b_{kj} Tv_k = \sum_{k=1}^n b_{kj} \bigl( \sum_{i=1}^m a_{ik} w_i \bigr)\\ &= \sum_{i=1}^m \bigl(\sum_{k=1}^n a_{ik} b_{kj} \bigr) w_i. \end{split} \end{equation*}$$

Hence, the matrix $C=(c_{ij})$ is given by
$$\begin{equation} \label{eq:c} c_{ij} = \sum_{k=1}^n a_{ik} b_{kj}. \tag{6.6.2} \end{equation}$$

Equation 6.6.2 can be used to define the $m\times p$ matrix $C$ as the product of a $m\times n$ matrix $A$ and a $n\times p$ matrix $B$, i.e.,
$$\begin{equation} C = AB. \tag{6.6.3} \end{equation}$$

Our derivation implies that the correspondence between linear maps and matrices respects the product structure.

Proposition 6.6.5. Let $S:U\to V$ and $T:V\to W$ be linear maps. Then

$$M(TS) = M(T)M(S).$$

Example 6.6.6. With notation as in Examples 6.6.3 and 6.6.4, you should be able to verify that
$$\begin{equation*} M(TS) = M(T)M(S) = \begin{bmatrix} 2&1\\ 5&2 \\ 3&1 \end{bmatrix} \begin{bmatrix} 2&1\\- 3& -2 \end{bmatrix} = \begin{bmatrix} 1&0\\ 4&1\\ 3&1 \end{bmatrix}. \end{equation*}$$

Given a vector $v\in V$, we can also associate a matrix $M(v)$ to $v$ as follows. Let $(v_1,\ldots,v_n)$ be a basis of $V$. Then there are unique scalars $b_1,\ldots,b_n$ such that

$$v= b_1 v_1 + \cdots b_n v_n.$$

The matrix of $v$ is then defined to be the $n\times 1$ matrix

$$M(v) = \begin{bmatrix} b_1 \\ \vdots \\ b_n \end{bmatrix}.$$

Example 6.6.7 The matrix of a vector $x=(x_1,\ldots,x_n) \in \mathbb{F}^n$ in the standard basis $(e_1,\ldots,e_n)$ is the column vector or $n \times 1$ matrix
$$\begin{equation*} M(x) = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} \end{equation*}$$
since $x=(x_1,\ldots,x_n) = x_1 e_1 + \cdots + x_n e_n$.

The next result shows how the notion of a matrix of a linear map $T:V\to W$ and the matrix of a vector $v\in V$ fit together.

Proposition 6.6.8. Let $T:V\to W$ be a linear map. Then, for every $v\in V$,
$$\begin{equation*} M(Tv) = M(T) M(v). \end{equation*}$$

Proof.

Let $(v_1,\ldots,v_n)$ be a basis of $V$ and $(w_1,\ldots,w_m)$ be a basis for $W$. Suppose that, with respect to these bases, the matrix of $T$ is $M(T)=(a_{ij})_{1\le i\le m, 1\le j\le n}$. This means that, for all $j\in \{1,2,\ldots,n\}$,

$$\begin{equation*} Tv_j = \sum_{k=1}^m a_{kj} w_k. \end{equation*}$$

The vector $v\in V$ can be written uniquely as a linear combination of the basis vectors as

$$v = b_1 v_1 + \cdots + b_n v_n.$$

Hence,

$$\begin{equation*} \begin{split} Tv &= b_1 T v_1 + \cdots + b_n T v_n\\ &= b_1 \sum_{k=1}^m a_{k1} w_k + \cdots + b_n \sum_{k=1}^m a_{kn} w_k\\ &= \sum_{k=1}^m (a_{k1} b_1 + \cdots + a_{kn} b_n) w_k. \end{split} \end{equation*}$$

This shows that $M(Tv)$ is the $m\times 1$ matrix

$$\begin{equation*} M(Tv) = \begin{bmatrix} a_{11}b_1 + \cdots + a_{1n} b_n \\ \vdots \\ a_{m1}b_1 + \cdots + a_{mn} b_n \end{bmatrix}. \end{equation*}$$

It is not hard to check, using the formula for matrix multiplication, that $M(T)M(v)$ gives the same result.

Example 6.6.9. Take the linear map $S$ from Example 6.6.4 with basis $((1,2),(0,1))$ of $\mathbb{R}^2$. To determine the action on the vector $v=(1,4)\in \mathbb{R}^2$, note that $v=(1,4)=1(1,2)+2(0,1)$. Hence,
$$\begin{equation*} M(Sv) = M(S)M(v) = \begin{bmatrix} 2&1\\-3&-2 \end{bmatrix} \begin{bmatrix} 1\\2 \end{bmatrix} = \begin{bmatrix} 4\\ -7 \end{bmatrix}. \end{equation*}$$

This means that

$$Sv= 4(1,2)-7(0,1)=(4,1),$$

which is indeed true.

Contributors

• Isaiah Lankham, Mathematics Department at UC Davis
• Bruno Nachtergaele, Mathematics Department at UC Davis
• Anne Schilling, Mathematics Department at UC Davis

Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.