Is the matrix of a Hermitian operator on a finite dimensional vector space always Hermitian regardless of the basis?

Question

I'm trying to prove that if an operator $U$ is Hermitian, then its corresponding matrix in an arbitrary basis $U_{ij}$ satisfies $U_{ij}^* = U_{ji}$, but it seems like that's only possible if the basis is orthonormal. Is this false?

Answer 1

Let $V$ be a complex inner product space and $f\colon V\to V$ a Hermitian operator, that is $$\langle f(v),w\rangle = \langle v,f(w)\rangle.$$

Given a basis $B=\{v_1,\dots,v_n\}$ of $V$ and denoting by $[v]_B\in\mathbb C^n$ the coordinate vector of $v\in V$, the operator $f$ corresponds to the matrix $$ [f]_B = \left.\bigg(\,[f(v_1)]_B\ \middle|\ \cdots\ \middle|\ [f(v_n)]_B\,\right). $$ Denote by $G_B=(\langle v_i, v_j\rangle)_{i,j}$ the Gramian matrix of the inner product with respect to the basis $B$, then $$ \langle v,w \rangle = [v]_B^\dagger \,G_B\, [w]_B, $$ where $M^\dagger$ denotes the conjugate transpose of a complex matrix $M$.

Hence, the fact that $f$ is Hermitian translates to $$ [f(v)]_B^\dagger\, G_B\, [w]_B = [v]_B^\dagger\, G_B\, [f(w)]_B, $$ which is equivalent to $$ [v]_B^\dagger\, [f]_B^\dagger\, G_B\, [w]_B = [v]_B^\dagger\, G_B\, [f]_B\, [w]_B. $$ Letting $v$ and $w$ run through the basis vectors of $B$, this becomes $$ [f]_B^\dagger\, G_B = G_B\, [f]_B. $$

If $B$ is an orthonormal basis, the Gram matrix $G_B$ is the identity matrix and this condition becomes $[f]_B^\dagger = [f]_B$. So in this case $f$ is Hermitian if and only if $[f]_B$ is a Hermitian matrix.

If $B$ is not an orthonormal basis, the Gramian matrix $G_B$ is just any positive definite Hermitian matrix and the condition doesn't simplify. Hermitian operators do not correspond to Hermitian matrices any longer.