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Let $\rho : \mathbb Z \to \mathrm{GL}_2(\mathbb C)$ be the representation defined by $\rho(1) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. I'd like to show that $\rho$ is not completely reducible.

i) I have one preliminary question (which is probably a silly one) - for what vector space $V$ is $\mathrm{GL}(V) \cong \mathrm{GL}_2(\mathbb C)$?

Firstly, I noted that $\rho(1)$ has an eigenvector, so the representation is not irreducible. So if it were completely reducible, it would have to break up as a direct sum of two 1-dimensional sub representations. But a 1-dimensional subrep is given by an eigenvector - but $\rho$ only has one eigenvalue, which has a 1-dimensional eigenspace. So this can't happen.

ii) Is this reasoning OK?

Once I've shown that the representation isn't irreducible, the problem is equivalent to showing that $\rho(1)$ cannot be diagonalised (which I've done by showing that the sum of the dimensions of the eigenspaces is 1, not 2).

Dependant on the answer to question i), I could have reduced (excuse the pun) the amount of work by considering Jordan Normal Form ($\rho(1)$ is in JNF but isn't diagonal, so isn't diagonalisable).

Thanks.

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    I think I *am* being silly with question i). $V = \mathbb C^2$ as a $\mathbb C$-vector space, right?2012-03-17
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    So the Jordan Normal Form argument *does* apply2012-03-17
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    Indeed, the vector space is $\mathbb C^2$., and yes, the JNF argument does work, too.2012-03-17
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    @Matt : The only such $V$'s you will find are those who are isomorphic to $\mathbb C^2$. This is the classical example where we have a counter-example to justify representation theory working only with finite groups. When in infinite group, the whole idea of computing characters to work with the representation is pointless.2012-03-17
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    @Patrick: Well... *pointless* is a bit too strong :) For example, in the correct context, the finite dimensional representation theory of many infinite groups (like semisimple Lie groups, say) is completely controled by characters.2012-03-17
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    Ohh. But then you're dealing with infinite character tables or something? Or does Maschke's theorem hold for those groups and the number of conjugacy classes is finite?2012-03-17
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    For simplicity, suppose you have a compact Lie group $G$. Then it has maximal compact subgroups, which are tori, and if we fix one, call it T, every element in G is conjugate to an element of T, so characters are determined by their restriction to T. Since T is not too complicated, one can do wonders with that.2012-03-17
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    Firstly, I noted that $\rho(1)$ has an eigenvector, so the representation is not irreducible.... Is this any proposition... if yes then can you give me its reference..thnx in advance @Matt2016-05-27

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