# 6.6: The matrix of a linear map

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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.