Let $T:V\rightarrow V$ bwe a linear transformation. Let $L \subset V$ be a linear subspace such that $L \cap \text{Ker}\,(T)=\{0 \}$.
Prove that the image given by T of any linear independent sequence in L is linear independent sequence.
Actually, I didn't understand the question.
Please help (with understanding the question and solution), thank you!
Assume the contrary, i.e., that for $w_1,\ldots,w_k\in L$ which are linearly independent, their images $Tw_1,\ldots,Tw_k$ are linearly dependent, which means that there exist $c_1,\ldots,c_k$, not all zero, such that $$ 0=c_1Tw_1+\cdots+c_kTw_k=T(c_1w_1+\cdots+c_1w_k). $$ But this implies that $$ c_1w_1+\cdots+c_1w_k \in\mathrm{ker}\,T, $$ and as $$ c_1w_1+\cdots+c_1w_k \in L, $$ then $$ c_1w_1+\cdots+c_1w_k \in L\cap\mathrm{ker}\,T=\{0\}, $$ and hence $$ c_1w_1+\cdots+c_1w_k=0. $$ But since $w_1,\ldots, w_k$ are linearly independent, then $c_1=\cdots=c_k=0$, which is a contradiction.