As \(L\) is invertible, there exists \(M\from W\to V\) such that \(M\Comp L=\IdMapOn{V}\) and \(L\Comp M=\IdMapOn{W}\). Now let's turn to the proof of the two claims.

i.   \(\Vect{v}_1, \dots, \Vect{v}_r\) linearly independent in \(V\) implies \(L(\Vect{v}_1),\dots ,L(\Vect{v}_r)\) linearly independent in \(W\)

Suppose we have a linear combination

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

To show that the vectors \(L(\Vect{v}_1),\dots ,L(\Vect{v}_r)\) are linearly independent, we must show that \(t_1,\dots ,t_r=0\). We need to use that the vectors \(\Vect{v}_1,\dots ,\Vect{v}_r\) are linearly independent, and to be able to do so, we use \(M\) to reverse the transformation effect of \(L\):

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

... and here we have it: this is only possible if \(t_1,\dots ,t_r=0\) because the vectors \(\Vect{v}_1,\dots ,\Vect{v}_r\) are linearly in dependent.

ii.   If \(\Vect{a}_1,\dots ,\Vect{a}_s\) span \(V\), then \(L(\Vect{a}_1),\dots ,L(\Vect{a}_s)\) span \(W\)

Let \(\Vect{w}\in W\) be arbitrary. We must show that it can be expressed as a linear combination of the vectors \(L(\Vect{a}_1),\dots , L(\Vect{a}_s)\). So, let's see: \(M(\Vect{w})\in V\) can be expressed as a linear combination of \(\Vect{a}_1,\dots ,\Vect{a}_s\):

\[M(\Vect{w}) = t_1\cdot \Vect{a}_1 + \cdots + t_s\cdot \Vect{a}_s\]

Now, we use \(L\) to reverse the transformation effect of \(M\):

\(\Vect{w}\)	\(=\)	\(L\Comp M(\Vect{w})\)
\(\)	\(=\)	\(L\left( t_1\cdot \Vect{a}_1 + \cdots + t_s\cdot \Vect{a}_s\right)\)
\(\)	\(= \)	\(t_1\cdot L(\Vect{a}_1) + \cdots + t_s\cdot L(\Vect{a}_s)\)

This is what we wanted: \(\Vect{w}\) is a linear combination of \(L(\Vect{a}_1),\dots ,L(\Vect{a}_s)\). As \(\Vect{w}\) is arbitrary, this shows that the vectors \(L(\Vect{a}_1),\dots ,L(\Vect{a}_s)\) span \(W\).