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Let $V$ and $W$ be two vector spaces over $\mathbb{R}$ Suppose $X \subseteq V$ is a nonempty linearly independent set and $T:V \rightarrow W$ is an injective linear map. Prove that {$T(u): u \in X$} is linearly independent in $W$

My attempt:

Since $T: V \rightarrow W$ is an injective linear map, it follows that $Ker(T) = 0$. By the rank-nullity theorem, $Dim(W) = Rank(T) + Ker(T). $ Hence $Dim(W) = Rank(T)$

Suppose for some ($u_1...u_n) \in X \subseteq V$ and for some scalars $(c_1...c_n) \in \mathbb{R}$, that

$c_1T(u_1)+ ... + c_nT(u_n) = 0$

$\iff T(c_1u_1)+ ... + T(c_nu_n) = 0$

$\iff c_1u_1+ ... + c_nu_n = 0$

I'm not sure where to go from here, or if I'm even on the right track...

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    $\begingroup$ But is that really what rank-nullity says? $\endgroup$ – M10687 Jun 19 '16 at 0:27
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Your statement of the rank-nullity theorem is wrong but your second idea is good and leads to the proof. If $c_1 T(u_1) + \dots + c_n T(u_n) = 0$ then by linearity, $T(c_1 u_1 + \dots + c_n u_n) = 0$. Since $T$ is injective, we must have $c_1 u_1 + \dots c_n u_n = 0$ and since $X$ is linearly independent, we get $c_1 = \dots = c_n = 0$.

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