= Solution
Define the <bidual evaluation map> by evaluation:
$$
\boxed{J:V\longrightarrow V'',\qquad J(v)(\ell)=\ell(v).}
$$
For fixed $v$, this is a <linear functional> of $\ell$, so it belongs to the <double dual>. It is also linear as a function of $v$. If $v\ne0$, some coordinate of $v$ in a <basis> is nonzero, and the corresponding <dual basis> functional detects it. Hence $J(v)\ne0$, proving <injectivity>.
The spaces $V,V',V''$ all have the same finite <dimension>. Therefore an <injective> <linear map> $J:V\to V''$ is <surjective>, proving the required <linear isomorphism>.
This <linear isomorphism> is natural because evaluation requires no <basis>. More explicitly, for any <linear map> $A:V\to W$, its <dual map> satisfies
$$
\bigl(A''J_V(v)\bigr)(\ell)=J_V(v)(\ell\circ A)=\ell(Av)=J_W(Av)(\ell),
$$
so $A''J_V=J_WA$.
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