Properties of The Matrix of a Linear Map

# The Linear Map from The Set of Linear Maps to The Set of m x n Matrices

We have already looked at constructing The Matrix of a Linear Map given a linear transformation $T \in \mathcal L (V, W)$, basis $B_V$ of $V$, and a basis $B_W$ of $W$. We will now look at some important properties of these matrices. Before we do so, we will note that the notation $\mathrm{Mat} (m, n, \mathbb{F})$ represents the set of all $m \times n$ matrices whose entries come from the field $\mathbb{F}$.

 Theorem 1: If $T \in \mathcal L (V, W)$ where $V$ and $W$ are finite-dimensional vector spaces with bases $B_V = \{v_1, v_2, ..., v_n \}$ and $B_W = \{ w_1, w_2, ..., w_m \}$ respectively, then the function $\mathcal M : T \to \mathcal M (T, B_V, B_W)$ is a linear map from the vector space $\mathcal L (V, W)$ to $\mathrm{Mat} (m, n, \mathbb{F})$.
• Proof: To show that $\mathcal M : T \to \mathcal M (T, B_V, B_W)$ is a linear map, we will show that for $c, d \in \mathbb{F}$ and $S, T \in \mathcal L (V, W)$ that $\mathcal M (cS + dT) = c\mathcal M (S) + d \mathcal M (T)$. Let $\mathcal M (S) =\begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix}$ and let $\mathcal M (T) =\begin{bmatrix} b_{1,1} & b_{1,2} & \cdots & b_{1,n}\\ b_{2,1} & b_{2,2} & \cdots & b_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ b_{m,1} & b_{m,2} & \cdots & b_{m,n} \end{bmatrix}$. Therefore it follows that we have:
(1)
\begin{align} c \mathcal M (S) + d \mathcal M (T) = \begin{bmatrix} ca_{1,1} + db_{1,1} & ca_{1,2} + db_{1,2} & \cdots & ca_{1,k} + db_{1,k} & \cdots & ca_{1,n} + db_{1,n} \\ ca_{2,1} + db_{2,1} & ca_{2,2} + db_{2,2} & \cdots & ca_{2,k} + db_{2,k} & \cdots &ca_{2,n} + db_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ ca_{m,1} + db_{m,1} & ca_{m,2} + db_{m,2} & \cdots & ca_{m,k} + db_{m,k} & \cdots & ca_{m,n} + db_{m,n}\end{bmatrix} \end{align}
• In particular, the $k^{\mathrm{th}}$ column $c \mathcal M (S) + d \mathcal M (T)$ is $\begin{bmatrix} ca_{1,k} + db_{1,k}\\ ca_{2,k} + db_{2,k}\\ \vdots\\ ca_{m,k} + db_{m,k} \end{bmatrix}$.
• Now consider the following:
(2)
\begin{align} \quad (cS + dT)(v_k) = cS(v_k) + dT(v_k) \\ \quad (cS + dT)(v_k) = c(a_{1,k}w_1 + a_{2,k}w_2 + ... + a_{m,k}w_m) + d(b_{1,k}w_1 + b_{2,k}w_2 + ... + b_{m,k}w_m)\\ \quad (cS + dT)(v_k) = (ca_{1,k} + db_{1,k})w_1 + (ca_{2,k} + db_{2,k})w_2 + ... + (ca_{m,k} + db_{m,k})w_m \end{align}
• This implies that the $k^{\mathrm{th}}$ column of $\mathcal M (cS + dT)$ is $\begin{bmatrix} ca_{1,k} + db_{1,k}\\ ca_{2,k} + db_{2,k}\\ \vdots\\ ca_{m,k} + db_{m,k} \end{bmatrix}$. Therefore we conclude that $\mathcal M (cS + dT) = c\mathcal M (S) + d \mathcal M (T)$, and so $\mathcal M$ is a linear transformation. $\blacksquare$

# The Matrix of the Image of a Vector Under a Linear Map

 Theorem 2: If $T \in \mathcal L (V, W)$ and $V$ and $W$ are finite-dimensional $\mathbb{F}$ vector spaces with bases $B_{V} = \{ v_1, v_2, ..., v_n \}$ and $B_{W} = \{w_1, w_2, ..., w_m \}$ respectively, then $\forall v \in V$, $\mathcal M (T v)_{m \times 1} = \mathcal M (T)_{m \times n} \mathcal M (v)_{n \times 1}$.
• Proof: Let $\mathcal M (T) = \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,k} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,k} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{m,1} & a_{m,2} & \cdots & a_{m,k} & \cdots & a_{m,n} \end{bmatrix}$. Let $v \in V$, and so $v = b_1v_1 + b_2v_2 + ... + b_nv_n$, so $\mathcal M (v) = \begin{bmatrix} b_1\\ b_2\\ \vdots\\ b_n \end{bmatrix}$. Therefore:
(3)
\begin{align} \mathcal M (T)_{m \times n} \mathcal M (v)_{n \times 1} = \begin{bmatrix} a_{1,1}b_1 + a_{1,2}b_2 + \cdots + a_{1,n}b_n\\ a_{2,1}b_1 + a_{2,2}b_2 + \cdots + a_{2,n}b_n\\ \vdots\\ a_{m,1}b_1 + a_{m,2}b_2 + \cdots + a_{m,n}b_n \end{bmatrix}_{m \times 1} \end{align}
• Now also, $T(v) = T(b_1v_1 + b_2v_2 + ... + b_nv_n) = b_1T(v_1) + b_2T(v_2) + ... + b_nT(v_n) = b_1 \sum_{i=1}^{m} a_{i, 1}w_i + b_2\sum_{i=1}^{m} a_{i, 2}w_i + ... + b_n \sum_{i=1}^{m} a_{i, n}w_i$, and therefore we have that once again $\mathcal M (T v ) = \begin{bmatrix} a_{1,1}b_1 + a_{1,2}b_2 + \cdots + a_{1,n}b_n\\ a_{2,1}b_1 + a_{2,2}b_2 + \cdots + a_{2,n}b_n\\ \vdots\\ a_{m,1}b_1 + a_{m,2}b_2 + \cdots + a_{m,n}b_n \end{bmatrix}_{m \times 1}$.
• Therefore $\mathcal M (T v) = \mathcal M (T) \mathcal M (v)$. $\blacksquare$

# The Matrix of a Composition of Linear Maps

 Theorem 3: If $S \in \mathcal L (U, V)$ and $T \in \mathcal (V, W)$ where $U$, $V$, and $W$ are finite-dimensional $\mathbb{F}$-vector spaces with bases $B_{U} = \{u_1, u_2, ..., u_p \}$, $B_{V} = \{ v_1, v_2, ..., v_n \}$ and $B_{W} = \{ w_1, w_2, ..., w_m \}$ respectively, then $\mathcal M (T \circ S)_{m \times p} = \mathcal M (T)_{m \times n} \mathcal M (S)_{n \times p}$.