dedica66em
2022-01-05
Answered

Let V and W be vector spaces and $T:V\to W$ be linear. Let $\{{y}_{1},\dots ,{y}_{k}\}$ be a linearly independent subset of $R\left(T\right)$ . If $S=\{{x}_{1},\dots ,{x}_{k}\}$ 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

Answered 2022-01-06
Author has **28** answers

Let us take $\{{w}_{1},{w}_{2},{w}_{3}\dots {w}_{k}\}$ be the linear independent subset of R(t) instead of $\{{y}_{1}\dots {y}_{k}\}$ and $S=\{{v}_{1},{v}_{2},\dots {v}_{k}\}$ 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$S=\{{v}_{1},{v}_{2},\dots {v}_{k}\}$ 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,

$T({a}_{1}{v}_{1}+{a}_{2}{v}_{2}+\dots +{a}_{k}{v}_{k})=T\left(0\right)$ .

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$\{{w}_{1},{w}_{2},{w}_{3}\dots {w}_{k}\}$ is LI, the only choice is ${a}_{1}={a}_{2}\dots ={a}_{k}=0$ ...(3)

So, from equation (2) and (3), the set$S=\{{v}_{1},{v}_{2},\dots {v}_{k}\}$ is LI.

Hence, proved.

The transformation is linear, where V and W are two vector spaces.

To show

For any scalars

Since T is a function, the above equation can be written as follows,

Since T is linear transformation,

Since, the set

So, from equation (2) and (3), the set

Hence, proved.

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