I might need help here in understanding my own question in places and please don't hesitate in asking for edits and clarifications.
Background: A representation $\rho$ of a finite group $G$ is a group homomorphism from $G$ into $GL(V)$ for some vector space $V$.
If $W$ is a subspace of $V$ invariant under $\rho(G)$, then $\rho_{\left.\right|W}$ is called a subrepresentation. In the usual way we can show that every representation is a direct sum of irreducible representations.
If we endow $V$ with an inner product $\langle \cdot,\cdot\rangle$, then we can show that
$$\langle u,v\rangle_\rho=\sum_{t\in G}\langle \rho(t)u,\rho(t)v\rangle$$
is another and furthermore, with respect this inner product, the operators $\rho(s)$ are unitary.
Where $d_i$ is the dimension of the vector space $V_i$, where $\rho_i:G\rightarrow GL(V_i)$ it can be shown that the regular representation, which acts on the vector space $V_r=\mathbb{C}^{|G|}$, can be decomposed as
$$V_r=d_1V_1\oplus d_2V_2\oplus\cdots d_nV_n,$$
where $\{\rho_i\}_{i\in[n]}$ are the unitary irreducible representations
$\rho_i:G\rightarrow GL(V_i)$ and so $$r:G\rightarrow GL\left(\bigoplus_{i\in[n]}d_iV_i\right),$$
and we write
$$r=\bigoplus_{i\in[n]}d_i\rho_i.$$
Using the fact that the irreducible representations are equivalent to unitary ones allows us to show that the matrix elements of the unitary irreducible representations are orthogonal as elements of $F(G)$ with respect to the inner product
$$\langle f,g\rangle=\frac{1}{|G|}\sum_{t\in G}\overline{f(t)}\cdot g(t),$$
and as there are $|G|$ of them, the matrix elements form an orthogonal basis of $F(G)$.
Questions
- With respect to the canonical basis of $F(G)$, in what sense can I talk about the matrix elements of the unitary irreducible representations?
- Is there any natural way that the matrix elements of unitary irreducible representations form a basis?
Example: I think I have for $G=\mathbb{Z}_3$, that there are unitary irreducible representations $\{\tau,\rho_1,\rho_2\}$ with matrix elements
$$a_0:=\left(\begin{array}{c}1 \\ 1 \\ 1\end{array}\right),\,a_1:=\left(\begin{array}{c}1 \\ \omega \\ \omega_2\end{array}\right) \text{ and }\,a_2:=\left(\begin{array}{c}1 \\ \omega^2 \\ \omega\end{array}\right)\in F(\mathbb{Z_3}).$$
These are written with respect to the canonical basis of $F(\mathbb{Z_3})$ (although everything here is easier as $\mathbb{Z}_3$ is abelian).
Consider, with respect to the canonical basis of $F(G)$,
$$f=\left(\begin{array}{c}1 \\ 2 \\ 3\end{array}\right).$$
With respect to the basis $\{a_0,a_1,a_2\}$ basis, I think
$$f=\left(\begin{array}{c}2 \\ \frac{1}{\sqrt{3}}e^{5\pi i/6} \\ \frac{1}{\sqrt{3}}e^{7\pi i/6}\end{array}\right).$$
Best Answer
I just had this same question, and I think the answer is NO.
Let me first restate the question, since nobody seems to have understood how to phrase it.
Let $\rho: G \to C^{nxn}$ be a representation. All equivalent representations are given by $M^{-1}\rho M$ for invertible matrices $M$. The "Weyl unitary trick" shows that for any $\rho$, there is an equivalent unitary representation. The question is: is this unique?
The answer is NO. Suppose $\rho$ is a unitary representation. Let U be a unitary matrix. Consider $U^* \rho U$. This is also unitary: $$ (U^* \rho(g) U) (U^* \rho^*(g) U) = U^* \rho(g) \rho^*(g) U = U^* I U = I.$$
I hope this helps (assuming I understood the question correctly).