# Let V and W be vector spaces and T:V \rightarrow

Let V and W be vector spaces and $T:V\to W$ be linear. Let $\left\{{y}_{1},\dots ,{y}_{k}\right\}$ be a linearly independent subset of $R\left(T\right)$. If $S=\left\{{x}_{1},\dots ,{x}_{k}\right\}$ is chosen so that $T\left({}_{\xi }\right)={y}_{i}$ for $i=1,\dots ,k$, prove that S is linearly independent.
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Thomas Lynn
Let us take $\left\{{w}_{1},{w}_{2},{w}_{3}\dots {w}_{k}\right\}$ be the linear independent subset of R(t) instead of $\left\{{y}_{1}\dots {y}_{k}\right\}$ and $S=\left\{{v}_{1},{v}_{2},\dots {v}_{k}\right\}$ is subset of V such that $T\left({v}_{i}\right)={W}_{i}$...(1).
The transformation is linear, where V and W are two vector spaces.
To show $S=\left\{{v}_{1},{v}_{2},\dots {v}_{k}\right\}$ is LI.
For any scalars ${a}_{1},{a}_{2},\dots {a}_{k}$ consider ${a}_{1}{v}_{1}+{a}_{2}{v}_{1}+\dots +{a}_{k}{v}_{k}=0$...(2)
Since T is a function, the above equation can be written as follows,
$T\left({a}_{1}{v}_{1}+{a}_{2}{v}_{2}+\dots +{a}_{k}{v}_{k}\right)=T\left(0\right)$.
Since T is linear transformation,
$T\left({a}_{1}{v}_{1}\right)+T\left({a}_{2}{v}_{2}\right)+\dots +T\left({a}_{k}{v}_{k}\right)=0$
${a}_{1}T\left({v}_{1}\right)+{a}_{2}T\left({v}_{2}\right)+\dots +{a}_{k}T\left({v}_{k}\right)=0$
${a}_{1}{w}_{1}+{a}_{2}{w}_{2}+\dots +{a}_{k}{w}_{k}=0$ (From (1))
Since, the set $\left\{{w}_{1},{w}_{2},{w}_{3}\dots {w}_{k}\right\}$ is LI, the only choice is ${a}_{1}={a}_{2}\dots ={a}_{k}=0$...(3)
So, from equation (2) and (3), the set $S=\left\{{v}_{1},{v}_{2},\dots {v}_{k}\right\}$ is LI.
Hence, proved.