Theorem\(L\from V\to W\) isomorphism exists if and only if \(\mathit{Dim}(V)=\mathit{Dim}(W)\)

Let \(V\) and \(W\) be arbitrary subspaces of \(\RNrSpc{k}\). Then an isomorphism \(L\from V\to W\) exists if and only if \(\DimOf{V}=\DimOf{W}\).

Proof

Consider first the situation where there exists an isomorphism \(L\from V\to W\). If \(\mathcal{B}=\Set{\Vect{v}_1,\dots ,\Vect{v}_r}\) is any basis for \(V\), then we know two facts about the vectors \(L(\Vect{v}_1), \dots ,L(\Vect{v}_r)\):

  1. these vectors are linearly independent in \(W\) because they are linearly independent in \(V\), and
  2. these vectors span \(W\) because they span \(V\).

So, these vectors form a basis for \(W\); i.e. \(\DimOf{W}=r = \DimOf{V}\).

Consider next the situation where \(\DimOf{V}=\DimOf{W}\). Let us choose arbitary bases \(\mathcal{B}=\Set{\Vect{v}_1,\dots ,\Vect{v}_r}\) and \(\mathcal{C}=\Set{\Vect{w}_1,\dots ,\Vect{w}_r}\) for \(V\) and \(W\), respectively. Then linear functions are given by

\[L\from V \longrightarrow W,\quad L(\Vect{v}_1)\DefEq \Vect{w}_1,\dots ,L(\Vect{v}_r)\DefEq \Vect{w}_r\]

and

\[M\from W \longrightarrow V,\quad M(\Vect{w}_1)\DefEq \Vect{v}_1,\dots ,M(\Vect{w}_r)\DefEq \Vect{v}_r\]

Then

\[(M\Comp L)(\Vect{v}_1) = \Vect{v}_1,\ \dots\ , (M\Comp L)(\Vect{v}_r) = \Vect{v}_r\]

which means that \((M\Comp L) = \IdMapOn{V}\). Furthermore,

\[(L\Comp M)(\Vect{w}_1) = \Vect{w}_1,\ \dots\ , (L\Comp M)(\Vect{w}_r) = \Vect{w}_r\]

which means that \((L\Comp M) = \IdMapOn{W}\). This is exactly what we needed to show to prove that \(L\) and \(M\) are inverses of each other: i.e. \(L\from V\to W\) is invertible.