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.