In the book The Symmetric Group the author says:

Let $\chi$ and $\psi$ be characters of the $G$-module $V$.

By picking an orthonormal basis for $V$, we obtain a matrix representation $Y$ for $\psi$, where each $Y(g)$ is unitary; i.e.,

\begin{equation} Y(g^{-1}) = Y(g)^{-1} = \overline{Y(g)}^{t}. \end{equation}

How do we know that $Y$ is unitary with respect to every orthonormal basis?

I have also had a look at the these two posts:

  1. How to prove unitary matrices require orthonormal basis
  2. Unitary and transformation matrix

However, I cannot see why having an orthonormal basis for $V$ automatically implies that $Y$ is unitary. Clearly we cannot just write the columns of $Y$ by taking the basis vectors, as suggested in post 1 because this will probably not correspond to the transformation of $g \in G$ in $V$.

Thank you very much for your help!


If you have a representation $\varphi:G\rightarrow GL(V)$ of a finite group on a finite-dimensional complex vector space, then there is a procedure for building an inner product on $V$ with respect to which $G$ acts unitarily. Choose any Hermitian inner product $\langle,\rangle$ on $V$ and consider $\langle,\rangle_2$ defined by $$\langle v,w\rangle_2=\frac{1}{\vert G\vert}\sum_{g\in G}\langle gv,gw\rangle.$$ This is a $G$-invariant Hermitian inner product on $V$. So, if you take an orthonormal basis with respect to $\langle,\rangle_2$, then your matrix representatives will be unitary.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.