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There is some confusion about the distinction between self-adjoint and Hermitian matrices. Every answer I've come across posits that "a matrix is Hermitian if and only if it is self-adjoint w.r.t. the complex inner product". I'm chiefly concerned with finite matrices like you would encounter in any introductory linear-algebra textbook; no quantum-mechanical operators or infinite-dimensional shenanigans here. Just $A \in \mathbb{C}^{m\times m}$. To my understanding (correct me if I'm wrong),

  • Hermitian means $A = A^*$, the conjugate transpose.

  • Self-adjoint means $\langle A\vec x,\vec y\rangle = \langle\vec x,A\vec y\rangle$, where $\langle\vec x,\vec y\rangle = \vec x^*\vec y$. That is, for all matrices, there is an adjoint matrix $A^\dagger$ for which $\langle A^\dagger\vec x,\vec y\rangle = \langle\vec x,A\vec y\rangle$, but only for some matrices, $A = A^\dagger$: they are themselves their adjoint.

Proving that Hermitian matrices are self-adjoint is trivial. However, I can't find a proof of the converse. It makes intuitive sense: $\forall \vec v_1,\vec v_2\in\mathbb{C}^{m} : \langle \vec v_2,A\vec v_1 \rangle = \langle A\vec v_2, \vec v_1\rangle \iff \vec v_2^*A\vec v_1 = \vec v_2^*A^*\vec v_1\implies A = A^*$, because there are just too many choices of $\vec v_1$ and $\vec v_2$ for the middle equality to be a coincidence. I must either be missing something trivial here, or I lack the rigorous background to formalise my intuition.

Mew
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    Your idea about "too many choices for the equation to be a coincidence" is correct. The formal argument would be to choose $v_i=e_i =(0. \dots, 0,1,0, \dots, 0)$ (with the $1$ as the $i$-th component) and $v_j=e_j$ to get that the element $a_{ij}$ of both matrices $A$ and $A^$ are equal. Do that for every basis vector and you get $A=A^$. – Lukas Jun 16 '21 at 11:36
  • @Lukas That works! Consider posting it as an answer. – Mew Jun 16 '21 at 11:55

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