# Relationship between Unitary Matrices and Hermitian Inner Products

I'm working on proving whether a representation is unitarisable, so I need to show that for each matrix that my representations maps a group element to that $\langle \varphi (g) v | \varphi (g) w \rangle = \langle v | w \rangle$.

I know that given a matrix U which is unitary it holds that $\langle U v | U w \rangle = \langle v | w \rangle$. Where this is the standard Hermitian inner product. What I'm trying to figure out right now is how to get a matrix operator such that for a given inner product $\langle v | w \rangle _H$, we know that UU' = I implies that $\langle U v | U w \rangle _H = \langle v | w \rangle _H$

A comment on this post Unitary Matrices and the Hermitian Adjoint seems to say that the conjugate transpose has this implication for all inner products. I've tried to find a proof online of the equivalence but couldn't. So any help with that proof or insight into ways to prove that a representation isn't uniterisable would be very appreciated.

The problem is in the adjoint $U'$, as you denote it in your post. If $U'$ denotes the conjugate transpose, then the equivalence $$\tag{*} \langle Ux|Uy\rangle = \langle x|y\rangle\quad \iff \quad U' U=I$$ only holds if $\langle\cdot|\cdot\rangle$ is the standard inner product on $\mathbb C^n$. If you have the (pseudo-) inner product $$\langle x|y\rangle_A = \langle Ax| y\rangle,$$ where $A$ is Hermitian symmetric and nonsingular, then you should define the adjoint as follows: $$U^\star = A^{-1}U' A.$$ With this definition, you can check that $$\langle Ux|Uy\rangle_A = \langle x|y\rangle_A\quad \iff \quad U^\star U=I.$$
Note: I said "pseudo-inner product" because I never required $A$ to be positive definite. It can be or it can not be, and this has applications: those things are used in special relativity, where $A=\text{diag}(+1, -1,-1,-1)$.
• Yes, every inner product on $\mathbb C^n$ has that form. Given an inner product $(\cdot|\cdot)$, the matrix $A$ can be recovered by the formula $$a_{i{}j}=(e_i|e_j)$$ where $e_i$ is the standard basis of $\mathbb C^n$. Jun 25, 2017 at 22:11