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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!

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    If you don't mind me asking, what is the source of the cited proposition? – 2017-01-16
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    Proposition 9.5 in the book "Methods of representation Theory(I)" from Charles W. Curtis.. – 2017-01-16
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    It may be not easy to prove it, although, the authors said it is easy.. ....... – 2017-01-16
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    Linear maps between finite dimensional vector spaces of the same dimension are injective iff they are surjective. – 2017-01-16
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    you are right, thank you! – 2017-01-16

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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.