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Non-degenerate form induces adjoint linear operators


Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.9

Solution: We can always define an inner product on a finite-dimensional vector space $V$. Hence we shall assume that $V$ is a finite-dimensional inner product space.

As we have seen in previous exercises, Exercise 9.2.6 and Exercise 9.2.7, $T_f^*$ is invertible. We have\[f(S\alpha,\beta)=(T_fS\alpha|\beta)=(\alpha|S^*T_f^*\beta),\]\[f(\alpha,S’\beta)=(T_f\alpha|S’\beta)=(\alpha|T_f^*S’\beta).\]Hence it suffices to choose some $S’$ such that $S^*T_f^*=T_f^*S’$, that is $S’=(T_f^*)^{-1}S^*T_f^*$. Hence we showed the existence of $S’$.

Now we show the uniqueness. Suppose there exists another operator $S^\dagger$ such that $$f(S\alpha,\beta) = f(\alpha, S^\dagger\beta)$$ for all $\alpha$, $\beta$. Then $$f(\alpha, S^\dagger\beta)=f(\alpha, S’\beta),$$we have $$f(\alpha, (S’-S^\dagger)\beta)=0$$for all $\alpha$, $\beta$. Because $f$ is non-degenerate, we must have $(S’-S^\dagger)\beta=0$ for all $\beta$. Thus $S^\dagger=S’$. This shows the uniqueness.


Linearity

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