In Proposition, $A$ ia a $K$-algebra and $(A_A)^*=Hom_K(A_A,K)$. We want to get an an $A$-isomorphism $\theta$ from $_AA$ to $(A_A)^*$ by a nondegenerate bilinear $\beta$ by defining $\theta(b)(a)=\beta(a,b)$ for $a,b\in A$.
Why $\theta$ is an $A$-isomorphism from $_AA$ to $(A_A)^*$, when the bilinear $\beta$ is nondegenerate. $\theta$ is injective. However, how to prove that $\theta$ is surjective.
Thanks to everyone!
You have to use the assumption that $A$ is finite-dimensional over $K$ (otherwise, the statement is not true). Note that ${}_{A}A$ and $(A_A)^*$ have the same (finite) dimension over $K$, and $\theta$ is a $K$-linear injection between them. Since a linear map between vector spaces of the same dimension is injective iff it is surjective, $\theta$ is automatically also surjective.