 Solution

Define the bidual evaluation map by evaluation: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 $v\ne 0$, some coordinate of $v$ in a basis is nonzero, and the corresponding dual basis functional detects it. Hence $J(v)\ne 0$, proving injectivity.

The spaces $V,V^{\prime },V^{\prime \prime }$ all have the same finite dimension. Therefore an injective linear map $J:V\to V^{\prime \prime }$ 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 satisfiesso $A^{\prime \prime }{J}_V=J_{W}A$.