# Hermitian matrices and inner products

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 $$\overline{A}^T=A$$.

You have to choose an orthonormal basis, otherwise the statement is simply false. If you do so, then $$M=I_n$$.
• 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? Apr 24 at 8:38
• 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? Apr 24 at 8:46