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 \(L_{A^{-1}}\circ L_A=I\).