Nondegenerate symmetric bilinear forms

Let $K$ be field and $A$ is sub-algebra of $M_n(K)$. we define $b\colon A\times A \to K$ where $b(X,Y)=\operatorname{Tr}(XY)$. If $J(A )\not=0$, where $J(A)$ means Jacobson Radical of $A$, then $b$ is non-degenerate symmetric bilinear forms.

I can show $b$ is symmetric bilinear form, but I have no idea about non-degenerate.

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No,I don't think so.Why do you think $J(A)=0$ –  Babgen Apr 6 '12 at 17:37
I might be remembering something incorrectly. I'm sure you're right. –  Dylan Moreland Apr 6 '12 at 19:34