\((i) \implies (ii)\)   Suppose \(L\) is an isomorphism and \(L(\Vect{v})=\Vect{0}\). If \(\Vect{v}\neq \Vect{0}\), then the vector \(\Vect{v}\) by itself is linearly independent, hence \(L(\Vect{v}) \) is linearly independent. But then \(L(\Vect{v})\neq \Vect{0}\). Thus \(\Vect{0}\in V\) is the only vector for which \(L(\Vect{v}) = \Vect{0}\) is possible.

\((ii) \implies (iii)\)   Choose a basis \(\mathcal{B}=\Set{\Vect{v}_1,\dots ,\Vect{v}_r}\) for \(V\). We claim that \(\mathcal{C}\DefEq \Set{ L(\Vect{v}_1),\dots , L(\Vect{v}_c) }\) is a basis for \(W\). To see this, we begin by showing that \(\mathcal{C}\) is linearly independent. Indeed, starting from a linear combination of \(\Vect{0}\) in \(W\), we compute:

\(\Vect{0}\)	\(=\)	\(t_1\cdot L(\Vect{v}_1) + \cdots +t_r\cdot L(\Vect{v}_r)\)
\(\)	\(=\)	\(L(t_1\Vect{v}_1 +\cdots + t_r\Vect{v}_r)\)

By hypothesis, this means that \(t_1\Vect{v}_1+\cdots + t_r\Vect{v}_r = 0\). As \(\mathcal{B}\) is linearly independent, we have that \(\mathcal{C}\) is linearly independent. But then \(\mathcal{C}\) must also span \(W\); else \(W\) would have a basis with more than \(r\) vectors; a contradiction to the fact that \(\DimOf{W}=\DimOf{V}=r\). Consequence:

If \(\Vect{w}\in W\) is arbitrary, there exist \(t_1,\dots ,t_r\in \RNr\) such that

\[\Vect{w} = t_1\cdot L(\Vect{v}_1) + \cdots + t_rL(\Vect{v}_r) = L(t_1\cdot \Vect{v}_1 + \cdots + t_r\Vect{v}_r )\]

In other words, setting \(\Vect{v}\DefEq t_1\cdot \Vect{v}_1 + \cdots + t_r\Vect{v}_r\) yields \(L(\Vect{v})=\Vect{w}\), as required.

\((iii) \implies (i)\)   If \(\mathcal{B} = \Set{\Vect{v}_1,\dots, \Vect{v}_r}\) is a basis for \(V\), we claim that \(\mathcal{C}=\Set{ L(\Vect{v}_1),\dots ,L(\Vect{v}_r}\) is a basis for \(W\). Indeed, from assumption \((iii)\) we see first that \(\mathcal{C}\) spans \(W\). But then \(\mathcal{C}\) must also be linearly independent; else we could discard at least one vector in \(\mathcal{C}\), and the remainig vector would still span \(W\). But this would mean that the \(r\)-dimensional space \(W\) has a basis of fewer than \(r\) vectors; a contradiction.

So, now that we know that \(\mathcal{C}\) is a basis for \(W\), we can construct the following linear transformation:

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

By design, then, \(M\Comp L=\IdMapOn{V}\), and \(L\Comp M = \IdMapOn{W}\). In particular, this shows that \(L\) is an isomorphism.

This completes the proof.