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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.

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  • $\begingroup$ You could start by writing the condition that $S_v$ must satisfy in order to be linear. Then apply the definition of $S_v$. $\endgroup$
    – David Hill
    Nov 4 '17 at 0:30
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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 $$S_{v+\alpha v'}=S_v+\alpha S_{v'}.$$

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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.

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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. ..

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