For $x ∈ R^n$, define $λ_x : R^n → R$ by $λ_x(y) = x · y$
(a) Show that $λ_x ∈ Hom_R(R^n, R)$.
(b) Show that the map $x → λ_x$ establishes an isomorphism between $R^n$ and $Hom_R(R^n, R)$.
(c) Show that if {$v_1,v_2, . . . ,v_n$} is a basis for $R^n$, then {$λ_{v_1}, λ_{v_2} , . . . , λ_{v_n}$} is a basis for $Hom_R(R^n, R)$.
for part (a), I just showed that $λ_x$ was a linear transformation. Meaning it respects vector addition and scalar multiplication.
For part (b), I am a bit unsure of how to do this. I can see that the map is equivalent to $R^n → R$, and that is exactly what $Hom_R(R^n,R)$ does, but I don't know how to express it past an intuition. I thought of perhaps saying that both the given map and $Hom_R(R^n,R)$ can be expressed by $1$x$n$ matrix multiplication with a column vector in $R^n$, so I could perhaps conclude that it is of $dim(R^{1xn})$, or equivalently, $dim(R^n)$. Since the dimensions are the same, I can conclude $R^n$ is isomoprhic to $Hom_R(R^n,R)$. Does this work?
For part (c), I was just thinking that since $R^n$ is isomorphic to $Hom_R(R^n,R)$, I could say that each vector in the basis $v_i$ will be mapped to a unique element in $R$, corresponding to $λ_{v_i}$. Thus {$λ_{v_1}, λ_{v_2} , . . . , λ_{v_n}$} forms a basis for $Hom_R(R^n,R)$. But I'm not quite sure if I really showed what they wanted here.
Any help is greatly appreciated, thanks!