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Specifically, there is a passage in Dummit and Foote that says

Suppose $V$ is a finite-dimensional $FG$-module and $V$ is reducible. Let $U$ be a $G$-invariant subspace. Form a basis of $V$ by taking a basis of $U$ and enlarging it to a basis of $V$. Then for each $g \in G$ the matrix $\varphi(g)$, of $g$ acting on $V$ with respect to this basis is of the form

$\varphi(g) = [[\varphi_1(g); \psi(g)][0; \varphi_2(g)]]$, where $\varphi_1 = \varphi\vert_U$ (with respect to the chosen basis of $U$) and $\varphi_2$ is the representation of $G$ on $V/U$ (and $\psi$ is not necessarily a homomorphism - $\psi(g)$ need not be a square matrix). So reducible representations are those with a corresponding matrix representation whose matrices are in block upper triangular form.

How did they get that matrix?

Thanks!

($\varphi$ is the homomorphism from $G$ to $Aut(V)$, I'm not sure what $\psi$ is)

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1 Answer 1

up vote 2 down vote accepted

I think all they're saying is that if $U$ is invariant, then the lower left-hand block of the matrix must be $0$, since elements of $U$ only have entries in the upper part in this basis, and this must be invariant under multiplication by the matrix. Then $\psi$ and $\phi_2$ are just new names for the remaining blocks. This only makes sense if the notation $[[A;B][C;D]]$ stands for the matrix

$$ \left( \begin{array}{cc} A&B\\ C&D \end{array} \right)\;. $$

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Thank you! I think I need to review my linear algebra. –  badatmath May 28 '11 at 3:13

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