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I am self-studying from Linear Algebra by Hoffman and Kunze. In chapter 9, it is stated that every sesqui-linear form $f$ on a finite-dimensional inner product space $(V,(\cdot|\cdot))$ can be associated to a unique operator $T: V \to V$ such that $$f(\alpha,\beta) = (T\alpha|\beta), \quad \forall \alpha,\beta \in V.$$ It is then shown that $f$ is Hermitian if and only if $T$ is self-adjoint. I would like to know if there is there some property $\mathcal{P}$ of sesqui-linear forms such that $f$ satisfies $\mathcal{P}$ if and only if $T$ is normal.

I apologize if this is a trivial question, and I thank all of you in advance for the help.

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I'm afraid there is no such noteworthy property, otherwise it would have been already named. Nonetheless, let's recall that a normal operator satisfies the condition $T^\dagger T = TT^\dagger$, which can be reformulated as $(T\alpha,T\beta) = (T^\dagger\alpha,T^\dagger\beta)$. From there, it may be translated as $$ f(\alpha,T\beta) = (T\alpha,T\beta) = (T^\dagger\alpha,T^\dagger\beta) = (T^\dagger\beta,T^\dagger\alpha)^* = (T\beta,T\alpha)^* = f(\beta,T\alpha)^*, $$ which can be seen as an analog of the hermitian property (with an extra $T$).

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