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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!

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  • $\begingroup$ I assume your vector space has finite dimension. You are asked what happens, when you have $k$ linear independent vectors $v_1,v_2,...,v_k$, i.e. $\sum\limits{i=1,k}\lambda_iv_i=0$ only if $\lambda_i=0$ for all i, and then take a look at their images under T. There you have to show that their images are linear independent as well. Hint: Since $L \cap Ker(T)=\{0\}$, $T$ confined to $L$ is injective. $\endgroup$ – Maximilian M. Feb 18 '14 at 19:59
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    $\begingroup$ Wait, your math above says $L \cap Ker(T)=\{\theta \}$. With theta do you mean zero or some arbitrary vector called theta?? $\endgroup$ – Maximilian M. Feb 18 '14 at 20:06
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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.

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