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Form is non-degenerate if and only if the associated linear operator is non-singular

Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.6

Solution: If $f$ is non-degenerate, we show that $T_f$ is non-singular. Suppose $T_f\alpha=0$ for some $\alpha\in V$, then we have $$f(\alpha,\beta)=(T\alpha|\beta)=0$$ for all $\beta$. Since $f$ is non-degenerate, we must have $\alpha=0$. Hence $T_f$ is non-singular.

Conversely, suppose $f(\alpha,\beta)=0$ for all $\beta$, then we have $$f(\alpha,\beta)=(T_f\alpha|\beta)=0,$$for all $\beta$. In particular, we have $0=(T_f\alpha|T_f\alpha)$. Hence $T_f\alpha=0$. But $T_f$ is non-singular, therefore $\alpha=0$. Thus $f$ is non-degenerate.


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