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Let $L: \mathbb{R}^k \to \mathbb{R}^m$ be a linear transformation such that $\ker(L) = \{\overrightarrow{0} \}$. Prove that if $\{v_1,..., v_n\}$ is linearly independent set in $\mathbb{R}^k$ then $\{L(v_1), L(v_2), ... , L(v_n) \}$ is linearly independent.

We apply the definition. Consider scalars, $c_1, c_2, ... , c_n$, we must show that if

$c_1 L(v_1) + ... + c_n L(v_n) = 0$ then $c_1 = ... = c_n = 0$

by definition of linear mapping

$c_1L(v_1) + ... + c_n L(v_n) = L(c_1v_1 + ... + c_n v_n) = 0$, we know that $\ker(L) = \{\overrightarrow{0}\}$, thus

$c_1v_1 + ... + c_nv_n = 0$, and we are given $\{c_1, ... , c_n\}$ is LI, thus, $c_1 = ... = c_n = 0$, the proof is complete.

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    $\begingroup$ It is complete! $\endgroup$ – Arnaldo Apr 4 '17 at 21:45
  • $\begingroup$ I agree. Is there a question here? $\endgroup$ – Gregory Apr 4 '17 at 21:46
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    $\begingroup$ If your question is "does this look fine?" Then the answer is "yes, it does." $\endgroup$ – Pawel Apr 4 '17 at 21:46
  • $\begingroup$ Your proof is correct, perhaps consider your wording in the last sentence - the vectors $v_i$, rather than the scalars $c_i$, are linearly independent. $\endgroup$ – Dahn Apr 4 '17 at 21:47
  • $\begingroup$ Why are you not confident about your proof? If I want to add something to your proof, I would say that $\mathrm{ker}L=\{0\}$ implies $k\leq m$, which the question should have pointed out to clarify some absurdities like what if $n>m$. $\endgroup$ – taper Apr 5 '17 at 1:31

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