Left-multiplication transformation
Definition 1. Let $A \in M_{m \times n}(F)$. We denote by $L_A$ the mapping $L_A: F^n \longrightarrow F^m$ defined by $\mathsf{L}_A(x) = Ax, \forall x \in F^n$. We call $\mathsf{L}_A$ a left-multiplication transformation.
Theorem 1
Theorem 1. Let $A, B \in M_{m \times n}(F)$. Then we have the following properties:
(a) $L_A \in \mathcal{L}(F^n, F^m)$
(b) $[L_A]_{\beta}^{\gamma} = A$ where $\beta, \gamma$ are the standard ordered bases for $F^n, F^m$, respectively.
(c) $L_A = L_B \Longleftrightarrow A = B.$
(d) $L_{A+B} = L_A + L_B$ and $L_{aA} = aL_A, \forall a \in F$.
(e) If $T \in \mathcal{L}(F^n, F^m)$, then $! \exists C \in M_{m \times n}(F)$ such that $T = L_C$. In fact, $C = [T]_{\beta}^{\gamma}.$
(f) If $E \in M_{n \times p}(F)$, then $L_{AE} = L_AL_E.$
(g) If $m = n$, then $L_{I_n} = I_{F^n}.$
Proof.
(a) Let $x, y \in F^n, c \in F$. Then $L_A(cx + y) = A(cx + y) = cAx + Ay = cL_A(x) + L_A(y)$. Thus $L_A$ is linear.
(b) Note that $L_A(e_i) = Ae_i = [A]^i$ by Theorem 2. Thus $[L_A]_{\beta}^{\gamma} = A$.
(c) $\Longleftarrow$ is clear. Since $\forall x \in F^n, Ax = Bx$, $[A]^i = Ae_i = Be_i = [B]^i$ for each $i (1 \leq i \leq n)$. Thus $A = B$.
(e) Define $C := [T]_{\beta}^{\gamma}$. $\forall x \in F^n$, we have $[T(x)]_{\gamma} = [T]_{\beta}^{\gamma}[x]_{\beta} = C[x]_{\beta} = [L_C]_{\beta}^{\gamma}[x]_{\beta} = [L_C(x)]_{\gamma}$. Thus $T = L_C$. The uniqueness follows from (c).
(d), (f), (g) Trivial. $\blacksquare$
