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Proof That the Image of a Linearly Independent Set By a One to One Linear Transformation is Linearly Independent

Theorem
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.