Definition: Let $x\in V$ then define the map $\hat{x}$ $:$ $V^* \rightarrow \mathbb{F}$ to be such that $\hat{x}(f)=f(x)$.
Definition: Define $a_V : V \rightarrow V^{**}$ to be such that $a_V(v)=\hat{v}$.
If $T:V\rightarrow W$ is a linear map, its bidual map is denoted as $T^{**} : V^{**}\rightarrow W^{**}$
Remark: The above maps are both linear transformations, and $a_v$ is an isomorphism between the space and its bidual.
Let $T: V\rightarrow W$ be a linear map. Show that if $V,W$ are finite dimensions, then $T^{**}=T$
Observations: Let $\{$ $v_1,v_2....,v_n$ $\}$ and $\{$ $w_1,w_2....,w_m$ $\}$ be basis of $V$ and $W$ respectively. Observe that $T^{**} \circ a_V= a_W \circ T$ ( I have verified this). I know that I must define the unique linear map that is determined by the basis to show that they are equal, what must that map be?