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Here is a problem I faced in algebra. $\rho: A_4 \rightarrow End_{\mathbb C}\mathbb C^{10}$ is a representation of $A_4$. Then show that there is a vector $v \in \mathbb C^{10}$ such that $v$ is an eigenvector for all $\rho(g)$, $g \in A_4$. I think I miss some trick, or something. Any idea will be appreciated.

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The question is not clear. Are you trying to prove that if $\rho$ is a representation... then there is a vector $v$ ...? –  Gerry Myerson Aug 17 '13 at 6:37
    
@GerryMyerson you are right, I edited the question! –  user51128 Aug 17 '13 at 6:51
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up vote 3 down vote accepted

A vector $v \in \mathbb C^{10}$ is an eigenvector for all $\rho(g)$ if and only if $\mathbb Cv$ is a one dimensional invariant subspace. So you want to show that every representation of dimension $10$ must have a $1$ dimensional irreducible summand. If you compute the dimensions of the irreducible representations of $A_4$ you'll find that they are $1$, $1$, $1$, and $3$. As $3$ does not divide $10$ one of the summands of your representation must have dimension $1$.

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I was thinking about that, but how you know that the irreducible representations have 1, 1, 1 and 3 dimensions? I know two ways to figure it out, one is well-known equation of order to the sum of squares of dims, other is by number of conjugacy classes, but it seems that it is not easy task to compute the number of conjugacy classes explicitly. –  user51128 Aug 17 '13 at 7:30
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+1 @user51128: I would be very surprised, if the dimensions of the irreducible representations of $A_4$ (and $S_4$) had not been calculated in an example/exercise prior to asking this question in any textbook on representation theory. –  Jyrki Lahtonen Aug 17 '13 at 7:52
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