# Prove linearity of map involving dual space

For the problem below, I'm not yet used to proving linearity where dual spaces are involved so any help with this problem will be appreciated.

Let $$V$$ be a vector space over some field $$F$$, and $$V^*=\mathcal{L}(V,F)$$ its dual space. For every fixed $$v\in V$$, define a map

$$S_v:V^*\rightarrow F, T\mapsto T(v).$$

a) Show that $$S_v$$ is a linear map

b) Show that $$V\to (V^*)^*,\ v\mapsto S_v$$ is linear

c) Show that the map in part b is injective, and that for $$\dim V<\infty$$, it is an isomorphism.

• You could start by writing the condition that $S_v$ must satisfy in order to be linear. Then apply the definition of $S_v$. Nov 4 '17 at 0:30

For part $(a)$ you need to show that each $S_v$ is linear. That is, for each $T,T'\in V^*$ and each scalar $\alpha\in F$, we have $$S_v(T+\alpha T')=S_v(T)+\alpha S_v(T').$$
For part $(b)$ you need to show that the map $v\mapsto S_v$ is linear. That is, for each $v,v'\in V$ and each scalar $\alpha\in F$ we have have $$S_{v+\alpha v'}=S_v+\alpha S_{v'}.$$
For a) you need to check that $S_v(aT+bU) = aS_v(T)+bS_v(U)$. Remember that $aT+bU$ is defined pointwise: its value at $v$ is $(aT+bU)(v) = aT(v)+bU(v)$.
For b) you need to check that $S_{av+bu} = aS_v+bS_u$. Check that these two functionals agree on any argument $T$. On the left-hand side, use that $T$ is itself a linear map. Remember what $S_v$ does to linear maps.
To see part 3, note $S_v=0 \implies S_v(v^i)=0 ,\forall i$, where $v^1,\dots, v^n$ is the basis of dual one forms for $V^*$ with $v^i(v_j)=\delta _i^j$, ( the kronecker delta. .. ) where $v_1,\dots, v_n$ is a given basis for $V$... which implies that $v$ is the zero linear combination of the $v_i$, so zero. ..
Finally use the fact that, in the finite dimensional case, the dual space has the same dimension as $V$. We wind up with a linear injection between $n-$dimensional vector spaces, hence an isomorphism. ..