# Linear algebra/representation theory - an example needed

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: $$\Gamma_s = K^{-1} \cdot \Gamma_s' \cdot K \quad \forall s \in G.$$

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.

• Yes, @john, that's what I mean. – ancientmathematician Apr 17 at 10:28
• 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. – ancientmathematician Apr 17 at 10:36
• 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 $\kappa$? – johnny Apr 24 at 11:38
• This is basic linear algebra, surely you can diagonalise a matrix? – ancientmathematician Apr 24 at 15:58
• Perhaps if you swapped the eigenvectors it would help? – ancientmathematician Apr 25 at 14:04

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 $$y=2x$$.

• 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. – David Hill Apr 17 at 17:25