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This is an example given in the representation theory book:

Let $\rho : S_3 → GL_2(\mathbb C)\qquad$ be specified on the generators $(1 2)$ and $(1 2 3)$
by $\rho _{(1 2)} =$ $\begin{bmatrix} -1 & -1 \\ 0 & 1 \\ \end{bmatrix}$,
$\rho_{(1 2 3)} =\begin{bmatrix} -1 & -1 \\ 1 & 0 \\ \end{bmatrix}$
Claim: $\rho _{(1 2)}$ and $\rho _{(1 23)}$ do not have a common eigenvector.
Indeed, direct computation reveals $\rho _{(1 2)}$ has eigenvalues $1$ and $−1$ with $V_{−1} = \mathbb C e_1$ and $V_1 =\mathbb C \begin{bmatrix} -1 \\ 2 \\ \end{bmatrix}$

First of all, how do they compute $V_{−1}$ and $V_{1}$? Also, what do $V_{−1}$ and $V_{1}$ specify?

Clearly $e_1$ is not an eigenvector of $\rho _{(1 2 3)}$ as $\rho _{(1 2 3)}\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} =\begin{bmatrix} -1 \\ 1 \\ \end{bmatrix}$ . Also, $\rho _{(1 2 3)}\begin{bmatrix} -1 \\ 2 \\ \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \\ \end{bmatrix}$ , so (−1, 2) is not an eigenvector of $\rho _{(1 2 3)}$. Thus $\rho _{(1 2 )}$ and $\rho _{(1 2 3)}$ have no common eigenvector, which implies that $\rho$ is irreducible.

Secondly what is the reasoning made above while finding eigenvectors?
Also when I try to compute eigenvalues for $\rho _{(1 2 3)}$ , I get $\lambda^2+\lambda +1=0$ which does not have real roots

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  • $\begingroup$ Do you know how to compute eigenvectors? Because, they just computed them. Also, a matrix can have real valued eigenvectors even though the eigenvalues are complex. $\endgroup$ – Mathematician 42 Feb 24 '18 at 23:12
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The characteristic polynomial of $A=\begin{pmatrix}-1&-1\\0&1\end{pmatrix}$ is $\det\begin{pmatrix}-1-X&-1\\0&1-X\end{pmatrix}=-(1+X)(1-X)=X^2-1$. Hence the eigenvalues are $\pm 1$. The space $V_1$ is $V_1=N(A-1\cdot Id)$ by definition, that is the solutions to the system $(A-Id)X=0$. Easy calculation shows that $V_1=\text{Span}[(-1,2)]$. Similarly, $V_{-1}=\text{Span}[(1,0)]$.

Let $B=\begin{pmatrix} -1&1\\1&0 \end{pmatrix}$. Simple calculations show that neither $(-1,2)$ or $(1,0)$ are eigenvectors of $B$. Thus $A$ and $B$ share no eigenvectors. No need to compute the eigenvectors of $B$ explicitly!

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  • $\begingroup$ Oh,thanks! I didn't think that by " the notation $\mathbb C$ " they were actually intending to mean spanning. $\endgroup$ – Leyla Alkan Feb 24 '18 at 23:51
  • $\begingroup$ It's common notation, it is only used for one-dimensional spaces :) $\endgroup$ – Mathematician 42 Feb 24 '18 at 23:55

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