Let $\textsf{V}$ be a finite dimensional vector space over the field $F$. Show that $\textsf{T}\mapsto \textsf{T}^t$ is an isomorphism of $\mathcal{L}(\textsf{V},\textsf{V})$ onto $\mathcal{L}(\textsf{V}^*,\textsf{V}^*)$.
I have to show that a map from $\mathcal{L}(\textsf{V},\textsf{V})$ to $\mathcal{L}(\textsf{V}^*,\textsf{V}^*)$ defined by $h(\textsf{T}) = \textsf{T}^t$.
And to show $h$ is well defined homomorphism and one-to-one I am having difficulty in showing well defined.
Let $y\in V$ and $T\in L(V,V).$ Then, $T^*:V^*\to V^*$ is defined by $\langle T^*x^*,y\rangle=\langle x^*,Ty\rangle.$ If $\langle x^*,Ty\rangle=\langle x^*,T_1y\rangle$ for some other $T_1:V\to V,$ then $\langle x^*,(T_1-T)y\rangle=0.$ This last equality is true for all $x^*\in L(V,V)$ so $(T_1-T)y=0.$ And since this holds for all $y\in V$, it follows that $T_1=T.$