# When is the operator associated to a sesquilinear form normal?

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.

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 $$T$$).

• I understand. Thank you. Feb 27 at 7:14