
# 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
\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}

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(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
\label{eq:c}
c_{ij} = \sum_{k=1}^n a_{ik} b_{kj}. \tag{6.6.2}

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.,

C = AB. \tag{6.6.3}

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.

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