If
\[T\]
is a one to one linear transformation with domain \[D\]
and \[S \subset D\]
is linearly independent then \[T(S)\]
is also linearly independent.Proof
Suppose
\[\left\{ \mathbf{v_1} , ..., \mathbf{v_n} \right\}\]
be a linearly independent set of vectors so that \[\alpha_1 \mathbf{v_1} + ...+ \alpha_n \mathbf{v_n}\neq 0\]
for any scalars \[\alpha_1 , ..., \alpha_n\]
.Transforming by
\[T\]
gives\[\begin{equation} \begin{aligned} 0=T(0) & \neq T(\alpha_1 \mathbf{v_1} + ...+ \alpha_n \mathbf{v_n}) \\ &= T(\alpha_1 \mathbf{v_1}) + ...+ T(\alpha_n \mathbf{v_n})\\ &=\alpha_1 T(\mathbf{v_1}) + ...+ \alpha_n T(\mathbf{v_n}) \end{aligned} \end{equation}\]
\[T\]
is one to one hence \[T(x)=T(0)=0 \rightarrow x=0\]
so that \[\alpha_1 \mathbf{v_1} + ...+ \alpha_n \mathbf{v_n}\neq 0\]
, contadicting the assumption that \[\mathbf{v_1}, ..., \mathbf{v_n}\]
are linearly independent.