# Prove the relation T = S.

Prove the relation $T=S.$
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Bertha Stark
The vector space
Suppose vector v in V.
$v={a}_{1}{v}_{1}+{a}_{2}{v}_{2}+\dots +{a}_{n}{v}_{n}$
where, are scalars.
Suppose, $S\left(v\right)=S\left({a}_{1}{v}_{1}+{a}_{2}{v}_{2}+\dots +{a}_{n}{v}_{n}\right).\left(1\right)$
As S is a linear transformation therefore,
$S\left({v}_{1}+{v}_{2}\right)=S\left({v}_{1}\right)+S\left({v}_{2}\right)$
$S\left(av\right)=aS\left(v\right)$
$S\left(v\right)=S\left({a}_{1}{v}_{1}+{a}_{2}{v}_{2}+\dots +{a}_{n}{v}_{n}\right)$
$=S\left({a}_{1}{v}_{1}\right)+S\left({a}_{2}{v}_{2}\right)+\dots +S\left({a}_{n}{v}_{n}\right)$
$={a}_{1}S\left({v}_{1}\right)+{a}_{2}S\left({v}_{2}\right)+\dots +{a}_{n}S\left({v}_{n}\right)$
Also, $S\left({v}_{i}\right)=T\left({v}_{i}\right)$
So,
$S\left(v\right)={a}_{1}S\left({v}_{1}\right)+{a}_{2}S\left({v}_{2}\right)+\dots +{a}_{n}S\left({v}_{n}\right)$
$={a}_{1}T\left({v}_{1}\right)+{a}_{2}T\left({v}_{2}\right)+\dots +{a}_{n}T\left({v}_{n}\right)$
$=T\left({a}_{1}{v}_{1}+{a}_{2}{v}_{2}+\dots +{a}_{n}{v}_{n}\right)$
$=T\left(v\right)$
As, v in V is arbitrary.
$S\left(v\right)=T\left(v\right)$ for all v in V.
This proves that $S=T.$