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**Gamma** I assume you meant to add $\displaystyle CT(u)=T(Cu)$

This is just the dual space to $\displaystyle \mathbb{R}^n$

$\displaystyle (T+CS)(u)=T(u)+CS(u)=T(u)+S(Cu)$

Now all you gotta do is show it is actually a linear transformation itself, ie $\displaystyle (\mathbb{R}^n)^*$ showing its closed under scalar multiplication and addition.

$\displaystyle (T+CS) (au+bv)=T(au+bv)+S(C(au+bv))=aT(u)+bT(v)+CaS(u)+Cb S(v)$

$\displaystyle =aT(u)+aCS(u)+bT(v)+bCS(v) = a(T+CS)(u)+b(T+CS)(v)$

So it indeed a linear transformation of $\displaystyle \mathbb{R}^n$, QED.