Abstract We consider linear transformations \(L\from V\to W\) between subspaces of \(\RNrSpc{k}\) whose transformation effect can be reversed in the sense defined here.
Outline Recall: a linear function \(L\from \RNrSpc{n} \to \RNrSpc{n}\) is invertible if its transformation effect can be reversed. We now extend this property to linear transformations \(L\from \VSpc{V}\to \VSpc{W}\) between subvector spaces of \(\RNrSpc{k}\): We call \(\LinMap{L}\) invertible exactly when there exists a linear map \(\LinMap{M}\from \VSpc{W}\to \VSpc{V}\) such that \(\LinMap{L}\) and \(\LinMap{M}\) reverse each other's transformation effect.
Recognizing invertible linear transformations Next we look for criteria to identify invertible linear maps, and we find that \(\LinMap{L}\) is invertible if and only if the matrix representing \(\LinMap{L}\) with respect to any choice of ordered bases of \(\VSpc{V}\) and \(\VSpc{W}\) is invertible.
Further, we ask for conditions which are necessary so that an invertible linear transformation \(L\from V\to W\) can exist. We find that such \(L\) exists if and only if \(V\) and \(W\) have the same dimension. Indeed, we show that an invertible \(L\) transforms any basis of \(V\) into a basis of \(W\). So, \(\DimOf{V}=\DimOf{W}\).
On the other hand if the dimensions of \(V\) and \(W\) are equal, we can construct an invertible \(L\from V\to W\) by choosing bases \(\mathcal{B}=(\Vect{v}_1,\dots ,\Vect{v}_r)\) for \(V\) and \(\mathcal{C}=(\Vect{w}_1,\dots ,\Vect{w}_r)\) for \(W\) and constructing \(L\) as the linear map with \(L(\Vect{v}_1)=\Vect{w}_1\), ... , \(L(\Vect{v}_r)=\Vect{w}_r\). In particular, if \(\DimOf{V}=n\), then there is an isomorphism \(\RNrSpc{n}\to V\).
A linear map \(\LinMap{L}\from \VSpc{V}\to \VSpc{W}\) of subvector spaces of \(\RNrSpc{k}\) is called invertible or is an isomorphism if there exists a linear map \(\LinMap{M}\from \VSpc{W}\to \VSpc{V}\) such that
\[\LinMap{M}\Comp \LinMap{L} = \IdMapOn{V} \quad\text{and}\quad \LinMap{L}\Comp \LinMap{M} = \IdMapOn{W}\]In this case we call \(\LinMap{M}\) the inverse of \(\LinMap{L}\) and write \(\LinMap{M}=\LinMap{L}^{-1}\).
Invertible linear transformations are represented by invertible matrices:
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:
\(\LinMap{L}\) is invertible if and only if the representing matrix \(\LinMapMtrxxOnBss{L}{B}{A}\) is invertible.
If \(\LinMap{L}\) is invertible, then the matrix \(\LinMapMtrxxOnBss{L^{-1}}{A}{B}\) representing \(\LinMap{L^{-1}}\) satisfies
| \(\LinMapMtrxxOnBss{L^{-1}}{A}{B}\) | \(=\) | \(( \LinMapMtrxxOnBss{L}{B}{A} )^{-1}\) |
For an isomorphism \(L\from V\to W\) be an isomorphism of subspaces of \(\RNrSpc{k}\) the following hold:
If vectors \(\Vect{v}_1, \dots, \Vect{v}_r\in V\) are linearly independent, then \(L(\Vect{v}_1),\dots ,L(\Vect{v}_r)\in W\) are linearly independent.
If \(\Vect{a}_1,\dots ,\Vect{a}_s\) span \(V\), then \(L(\Vect{a}_1),\dots ,L(\Vect{a}_s)\) span \(W\).
Let \(V\) and \(W\) be arbitrary subspaces of \(\RNrSpc{k}\). Then an isomorphism \(L\from V\to W\) exists if and only if \(\DimOf{V}=\DimOf{W}\).
Let \(L\from V\to W\) be a linear transformation between subvector of \(\RNrSpc{k}\). If \(\DimOf{V} = \DimOf{W}\) then the following are equivalent:
\(L\) is an isomorphism
\(L(\Vect{v}) = \Vect{0}\) implies \(\Vect{v}=\Vect{0}\).
For every \(\Vect{w}\in W\) there exists \(\Vect{v}\in V\) such that \(L(\Vect{v}) = \Vect{w}\).
Let \(V\) be a subspace of \(\RNrSpc{k}\). If \(\DimOf{V}=n\), equipped with an ordered basis \(\mathcal{B}=(\Vect{b}_1,\dots ,\Vect{b}_n)\). Then the basis vector operation is linear and is isomorphism
\[M\from V \longrightarrow \RNrSpc{n},\qquad M(\Vect{x})\DefEq \CoordVect{\Vect{x}}{B}\]