TheoremInvertible linear map has invertible matrix

Let \(\LinMap{V}\) and \(\LinMap{W}\) be subspaces of \(\RNrSpc{k}\) with ordered bases \(\OrdVSpcBss{\EuScript{A}}\) and \(\OrdVSpcBss{\EuScript{B}}\), respectively. Then for a linear map \(\LinMap{L}\from \VSpc{V}\to \VSpc{W}\) the following hold:

  1. \(\LinMap{L}\) is invertible if and only if the representing matrix \(\LinMapMtrxxOnBss{L}{B}{A}\) is invertible.

  2. If \(\LinMap{L}\) is invertible, then the matrix \(\LinMapMtrxxOnBss{L^{-1}}{A}{B}\) representing \(\LinMap{L^{-1}}\) satisfies

    \(\LinMapMtrxxOnBss{L^{-1}}{A}{B}\)\(=\)\(( \LinMapMtrxxOnBss{L}{B}{A} )^{-1}\)

Proof

Consider first the case where \(\LinMap{L}\) is invertible; i.e. there is a linear map \(\LinMap{M}\from \VSpc{W}\to \VSpc{V}\) with

\[\LinMap{M}\Comp \LinMap{L} = \IdMapOn{V} \quad\text{and}\quad \LinMap{L}\Comp \LinMap{M} = \IdMapOn{W}\]

Let \(\LinMapMtrxxOnBss{L}{B}{A}\) and \(\LinMapMtrxxOnBss{M}{A}{B}\) denote the representing matrices. Then

\(\LinMapMtrxxOnBss{M}{A}{B}\cdot \LinMapMtrxxOnBss{L}{B}{A}\)\(=\)\(\LinMapMtrxxOnBss{M\Comp L}{A}{A} = \IdMtrx{}\)
\(\LinMapMtrxxOnBss{L}{B}{A}\cdot \LinMapMtrxxOnBss{M}{A}{B}\)\(=\)\(\LinMapMtrxxOnBss{L\Comp M}{B}{B} = \IdMtrx{}\)

This implies that both \(\LinMapMtrxxOnBss{L}{B}{A}\) and \(\LinMapMtrxxOnBss{M}{A}{B}\) are square shaped of the same size, and are inverse to each other; i.e.

\(\LinMapMtrxxOnBss{L^{-1}}{A}{B}\)\(=\)\(\left( \LinMapMtrxxOnBss{L}{B}{A}\right)^{-1}\)

Next consider the case where \(\Mtrx{A}\DefEq \LinMapMtrxxOnBss{L}{B}{A}\) is invertible. Then \(\Mtrx{B}\DefEq \Mtrx{A}^{-1}\) represents a linear function \(\LinMap{M}\from \VSpc{W}\to \VSpc{V}\). To see that \(\LinMap{M}\) does, indeed, reverse the transformation effect of \(\LinMap{L}\), we compute:

\(\CoordVect{(M\Comp L)(\Vect{x})}{\EuScript{A}}\)\(=\)\(\Mtrx{B}\Mtrx{A} \CoordVect{\Vect{x}}{\EuScript{A}} = \CoordVect{\Vect{x}}{\EuScript{A}}\)
\(\CoordVect{(L\Comp M)(\Vect{y})}{\EuScript{B}}\)\(=\)\(\Mtrx{A}\Mtrx{B} \CoordVect{\Vect{y}}{\EuScript{B}} = \CoordVect{\Vect{y}}{\EuScript{B}}\)

Thus \(\LinMap{L}\) is invertible, with inverse \(\LinMap{M}\). – This completes the proof.