The inverse of a linear mapping

Linear mappings: Linear mappings

The inverse of a linear mapping

Let $A$ be an $n\times n$ matrix. This defines a matrix mapping $L_A$ from $\mathbb{R}^n$ into $\mathbb{R}^n$ . We can apply this mapping repeatedly. We denote $(L_A)^2=L_A\circ L_A$ , $(L_A)^3=L_A\circ (L_A)^2$ , etc. Then: $(L_A)^n = L_{A^n}$

Let $A$ be an invertible $n\times n$ matrix. This defines a matrix mapping $L_A$ from $\mathbb{R}^n$ into $\mathbb{R}^n$ . This mapping is then invertible, say with inverse denoted by $(L_A)^{-1}$ . Then: $(L_A)^{-1} = L_{A^{-1}}$

We need to verify that $L_A\circ L_{A^{-1}} = L_{A^{-1}}\circ L_A=I$ : $\begin{array}{rcl}L_A\circ L_{A^{-1}} &=&L_{A\cdot A^{-1}}\\ &&\phantom{xxx}\blue{\text{composition of mappings determined by matrices}}\\ &=&L_{I}\\ &&\phantom{xxx}\blue{\text{definition of the inverse of a matrix}}\\ &=&I\\ &&\phantom{xxx}\blue{\text{multiplication of the identity matrix is the identity mapping}}\end{array}$ Similarly it can be proved that $L_{A^{-1}}\circ L_A=I$ .