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I am reading Linear representations of finite groups by Serre, the following makes sense but can I see this with a concrete example since I cannot think of one?

Let $\rho$ and $\rho\,'$ be two representations of the same group $G$ in vector spaces $V$ and $V\,'$. These representations are said to be isomorphic if there exits a linear homomorphism $\kappa\colon V \rightarrow V\,'$ which ''converts'' $\rho$ to $\rho\,'$, if the following equality holds $$ \kappa \circ \rho(s) = \rho(s)\,'\circ \kappa \quad \forall s \in G.$$

If $\rho$ and $\rho\,'$ are given in matrix form by $\Gamma_s$ and $\Gamma_s'$, this implies, there exists an invertible matrix $K$ such that the following is holds: $$ \Gamma_s = K^{-1} \cdot \Gamma_s' \cdot K \quad \forall s \in G. $$

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Let the base field be $\mathbb{C}$, and let $V=V'=\mathbb{V}_2(\mathbb{C})$.

Let $G=\mathbb{Z}_3=\langle d \rangle$.

In matrix terms let $$ \rho:d\mapsto \begin{pmatrix} 0 & 1\\-1 & -1\end{pmatrix} $$ and $$ \rho' :d\mapsto \begin{pmatrix}\omega & 0\\ 0 & \omega^2 \end{pmatrix} $$ where $\omega$ is the cube root of unity.

You can find the matrix of $\kappa$ for yourself, it's the usual matrix of eigenvectors.

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  • $\begingroup$ Yes, @john, that's what I mean. $\endgroup$ – ancientmathematician Apr 17 at 10:28
  • $\begingroup$ I found $\rho$ by writing down the companion matrix for an element of order $3$, which has to satisfy $(X^3-1)/(X-1)=X^2+X+1$. I know that the roots of $X^2+X+1$ are $\omega$ and $\omega^2$, so I write down $\rho'$. I know these will be similar because complex matrices with distinct eigenvalues are diagonalisable. I see that this explanation may not help. I think that before you tackle representation theory you should have a good grasp of linear algebra; there's a sense in which it is a gigantic generalisation of the theory of diagonalising matrices/decomposition into eigenspaces. $\endgroup$ – ancientmathematician Apr 17 at 10:36
  • $\begingroup$ now I understand, how you got the matrices for $\rho$ and $\rho'$ (I was being stupid aha!), so how can I get the matrix for $\kappa$? $\endgroup$ – johnny Apr 24 at 11:38
  • $\begingroup$ This is basic linear algebra, surely you can diagonalise a matrix? $\endgroup$ – ancientmathematician Apr 24 at 15:58
  • $\begingroup$ Perhaps if you swapped the eigenvectors it would help? $\endgroup$ – ancientmathematician Apr 25 at 14:04
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Take two representations of the cyclic group of order two $C_2=\langle c\mid c^2=1\rangle$ in $\mathbb{R}^2$ given by $$ \rho(c)=\begin{pmatrix}-1&0\\0&1\end{pmatrix} $$ and $$ \rho'(c)=\begin{pmatrix}\frac{3}{5}&-\frac{4}{5}\\-\frac{4}{5}&-\frac{3}{5}\end{pmatrix}. $$ The map $\kappa:\mathbb{R}^2\to\mathbb{R}^2$ is the map $$ \kappa\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix},\;\;\;\mbox{and}\;\;\;\kappa\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}-2\\1\end{pmatrix}. $$ Geometrically, $\rho$ represents $c$ as a reflection over the $y$-axis, and $\rho'$ represents $c$ as a reflection over the line $y=2x$.

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  • $\begingroup$ This is a basic linear algebra problem that you should be able to work out yourself. I told you what the matrices were: The first is a reflection over the line $x=0$ and the other is a reflection over the line $y=2x$. The linear map $\kappa$ implements a change of basis. $\endgroup$ – David Hill Apr 17 at 17:25

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