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Form is left non-degenerate if and only if it is right non-degenerate

Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.7


Call the form $f$ right non-degenerate if $0$ is the only vector $\alpha$ such that $f(\beta,\alpha)=0$ for all $\beta$.

Suppose $f$ is left non-degenerate, we show that $f$ is right non-degenerate. Suppose $f(\beta,\alpha)=0$ for all $\beta$. Hence $(T_f\beta|\alpha)=0$ for all $\beta$. Because $f$ is left non-degenerate,  $T_f$ is non-singular by Exercise 9.2.6. Since $V$ is finite-dimensional, we have $T_f$ is invertible. Hence there exists $\beta$ such that $T_f\beta=\alpha$. Therefore, we conclude that $$0=(T_f\beta|\alpha)=(\alpha|\alpha).$$Thus $\alpha=0$, namely $f$ is right non-degenerate.

Similar to Exercise 9.2.6, one can show that $f$ is right non-generate if and only if $T_f^*$ is non-singular. Then repeat our argument, we can show the other direction.


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