# Let V and W be vector spaces, let T : V \rightarrow W be linear, and let \{w_1 , w_2 , ...

Joyce Smith 2022-01-04 Answered
Let V and W be vector spaces, let $T:V\to W$ be linear, and let $\left\{{w}_{1},{w}_{2},\dots ,{w}_{k}\right\}$ be a linearly independent set of k vectors from R(T). Prove that if $S=\left\{{v}_{1},{v}_{2},...,{v}_{k}\right\}$ is chosen so that $T\left({v}_{i}\right)={W}_{i}$ for $i=1,2,\dots ,k,$ then S is linearly independent.
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## Expert Answer

Daniel Cormack
Answered 2022-01-05 Author has 34 answers

Definitions:
1) The set of vectors $\left\{{v}_{1},{v}_{2},...,{v}_{k}\right\}$ from a vector space V is said to be linearly independent if the linear combination of vectors ${\alpha }_{1}{v}_{1}+{\alpha }_{2}{v}_{2}+\dots +{\alpha }_{k}{v}_{k}=0$
$⇒{\alpha }_{1}={\alpha }_{2}=\dots ={\alpha }_{k}=0$ 2) Let V and W be vector spaces over the same field F.
The map $T:V\to W$ is said to be linear transformation from V to W if
(i) $T\left({0}_{v}\right)={0}_{w}$
(ii) $T\left(\alpha u+\beta v\right)=\alpha T\left(u\right)+\beta T\left(v\right)$ where $\alpha ,\beta \in F$ and $u,v\in V$
Let us consider an arbitrary representation of zero vector of V as a linear combination of vectors from S.
${0}_{v}={\alpha }_{1}{v}_{1}+{\alpha }_{2}{v}_{2}+\dots +{\alpha }_{k}{v}_{k}=\sum _{i=1}^{k}{\alpha }_{i}{v}_{i}$...(1) where ${\alpha }_{i}\in FS$ for $i=1,2,\dots ,k$
Since T is linear, ${0}_{v}\in V$ will always gets mapped to ${0}_{w}\in W$
i.e., $T\left({0}_{v}\right)={0}_{w}$
Thus, ${0}_{w}=T\left({0}_{v}\right)=T\left(\sum _{i=1}^{k}{\alpha }_{i}{v}_{i}\right)$
By the linearity of T we have
${0}_{w}=T\left(\sum _{i=1}^{k}{\alpha }_{i}{v}_{i}\right)=\sum _{i=1}^{k}{\alpha }_{i}T\left({v}_{i}\right)$
$\sum _{i=1}^{k}{\alpha }_{i}{w}_{i}$ (Since $T\left({v}_{i}\right)={w}_{i}$)
$={\alpha }_{1}{w}_{1}+{\alpha }_{2}{w}_{2}+\dots +{\alpha }_{k}{w}_{k}$
since the set $\left\{{w}_{1},{w}_{2},\dots ,{w}_{k}\right\}$ is linearly independent
$⇒{\alpha }_{1}={\alpha }_{2}=\dots ={\alpha }_{k}=0$
substituting ${\alpha }_{1}={\alpha }_{2}=...={\alpha }_{k}=0$ in (1) we have, the set $\left\{{v}_{1},{v}_{2},...,{v}_{k}\right\}$ are linearly independent.
Thus (1) will be the trivial representation of zero vector of V.
Hence $S=\left\{{v}_{1},{v}_{2},...,{v}_{k}\right\}$ is linearly independent.

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