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