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Let $V:=\mathbb{C}^n$ and $\langle \cdot, \cdot \rangle: V\times V\to \mathbb{C}$ be an inner product, which by definition is a positive-definite sesquilinear form. The adjoint of a linear operator $A: V\to V$ is denoted as $A^*$ and defined by $\langle Ax, y\rangle=\langle x, A^* y\rangle$, for all $x,y\in V$.

Why does a self-adjoint operator $A=A^*$ here correspond to an Hermitian matrix?

I know this is a basic fact. But I can't seem to prove it.

Let $B=\{e_1\cdots, e_n\}$ be a basis of $V$, and $\langle\cdot,\cdot\rangle$ anti-linear on the first argument. Then, the inner product has the associated matrix $M:=\left(\langle e_i, e_j\rangle\right)_{n\times n}$ such that $\langle x, y\rangle= \left\langle \sum_1^n c_x^i e_i, \sum_1^n c_y^j e_j\right\rangle=\overline{c_x}^T M c_y$. So, if $A=A^*$, and if denoting $A$ and $A^*$ also as their corresponding matrix forms w.r.t base $B$, we have $$\langle Ax,y\rangle=\overline{c_x}^T\overline{A}^TMc_y; \quad \langle x, Ay\rangle=\overline{c_x}^TMA c_y.$$ Then, by $\langle Ax,y\rangle=\langle x, A^*y\rangle=\langle x, Ay\rangle$ for all $x,y\in V$, we know $\overline{A}^T M= MA$. But I don't know how to conclude $\overline{A}^T=A$.

Thanks in advance.

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  • $\begingroup$ The basis has to be orthonormal for this to be true. $\endgroup$
    – Kandinskij
    Commented Apr 24, 2023 at 8:17

1 Answer 1

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You have to choose an orthonormal basis, otherwise the statement is simply false. If you do so, then $M=I_n$.

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  • $\begingroup$ Thanks! I originally thought about this as I'm trying to understand real symmetric matrices as linear operators on $\mathbb{C}^n$. It seems an operator $A$ corresponds to a real matrix iff there exists a basis $B=\{e_1,\cdots, e_n\}$ such that every $A(e_i)$ is a $\mathbb{R}$-linear combination of $B$. So $A$ must preserve $\text{span}_\mathbb{R}B$. Does that mean, for $A$ to have a real symmetric matrix form, the basis $B$ must also be orthonormal for a given inner product? $\endgroup$
    – user760
    Commented Apr 24, 2023 at 8:38
  • $\begingroup$ Yes, orthonormal basis are essential. $\endgroup$
    – Kandinskij
    Commented Apr 24, 2023 at 8:41
  • $\begingroup$ So if we know $A$ preserves $\text{span}_\mathbb{R}B$, and we know its matrix form w.r.t this basis is real and symmetric, does there always exists a compatible inner product? Can we just simply define it to satisfy $\langle e_i, e_j\rangle:=\delta_{ij}$, and then extend by sesquilinearity? Is $A$ only self-adjoint w.r.t this specific inner product? $\endgroup$
    – user760
    Commented Apr 24, 2023 at 8:46
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    $\begingroup$ Yes, that's true. $\endgroup$
    – Kandinskij
    Commented Apr 24, 2023 at 8:48
  • $\begingroup$ Thanks. I get it now. $\endgroup$
    – user760
    Commented Apr 24, 2023 at 8:50

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