Exercise 2.4.25

The transformation $\mathrm{\Psi }$ would be linear since

It will be injective by following arguments. If $\mathrm{\Psi }(f)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{f(s)\ne 0}f(s)s=0$ then we have that $f(s)=0$ on those $s$ such that $f(s)\ne 0$ since $\{s:f(s)\ne 0\}$ is finite subset of basis. But this can only be possible when $f=0$. On the other hand, we have for all element $x\in V$ we can write $x={\sum <!--\; FUNCTION\; APPLICATION\; -->}_i{a}_i{s}_i$ for some finite subset $\{s_i\}$ of $S$. Thus we may pick a function $f$ such that $f(s_i)=a_i$ for all $i$ and vanish outside. Thus $\mathrm{\Psi }$ will map $f$ to $x$. So $\mathrm{\Psi }$ is surjective. And thus it’s an isomorphism.