Composition of Linear Maps are not Commutative

Theorem

• $A,B\in {\mathbb{C}}^{m\times n}$
• Then, $\forall X\in C^{n\times 1},T_A{T}_B(x)=T_{AB}(x)$

Proof

• $T_A{T}_B(x)=T_A(Bx)=A(Bx)=(AB)(x)=T_{AB}(x)$
• As $T_{AB}\ne T_{BA}$ in general, it follows that compositions are not commutative $\square$