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Let $L : V → W$ be a one-to-one linear mapping. Show that if $B = \{v_1 , . . . , v_k \}$ linearly independent in $V$, then $\{L(v_1), . . . , L(v_k )\}$ is linearly independent in $W$.

Solution: Since $L$ is one-to-one we have that $\ker L =\{0\}$. Consider $$0 = c_1L(v_1 ) + · · · + c_nL(v_n)= L(c_1v_1 + · · · + c_nv_n )$$ This implies that $c_1 v_1 +· · ·+c_n v_n \in \ker( L)$. Thus, $c_1 v_1 +· · ·+c_n v_n = 0$. Thus, $c_1 = · · · = c_n = 0$ since $B$ is linearly independent. Therefore, $\{L(v_1 ), . . . , L(v_k )\}$ is linearly independent.


I don't understand why this proof rules out the possibility that there still could exist $k_1,...k_n$ such that $k_1L(v_1 ) + · · · + k_nL(v_n)$ is 0, just because $c_1...c_n$ is always 0.

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This question doesn't really make sense. The proof showed that for ARBITRARY $c_1,\cdots,c_n$ one has that $c_1L(v_1)+\cdots+c_nL(v_n)=0$ implies $c_1=\cdots=c_n=0$. This is precisely the statement that $\{L(v_1),\cdots,L(v_n)\}$ is linearly independent. Once again, they didn't pick some fixed $c_1,\cdots,c_n$ they showed the statement for ANY $c_1,\cdots,c_n$.

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the proofs says since $c_1...c_n =0$ then $c_1L(v_1)+⋯+c_nL(v_n)=0$ but why does this mean there doesn't exist a different solution? –  Mark Nov 12 '11 at 5:47
    
No, the proof says that given $c_1L(v_1)+\cdots+c_nL(v_n)=0$ THEN $c_1=\cdots=c_n=0$. You have your implications backwards. –  Alex Youcis Nov 12 '11 at 5:48
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Ok thanks, I see my misunderstanding –  Mark Nov 12 '11 at 5:50

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