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I am having a course about group representations and I saw this today:

If $T : V \rightarrow V$ is a linear transformation and $B$ is a basis for $V$ , then we shall use $[T]_B$ to denote the matrix for $T$ in the basis $B$. Let $\phi : G \rightarrow GL(V)$ be a decomposable representation, say with $V = V_1 \oplus V_2$ where $V_1$ and $V_2$ are non-trivial G-invariant subspaces. Let $\phi_i = \phi_{| V_i}$. Choose bases $B_1$ and $B_2$ for $V_1$ and $V_2$, respectively. Then it follows from the denition of a direct sum that $B$ = $B_1 \cup B_2$ is a basis for $V$ . Since $V_i$ is $G$-invariant, we have $\phi(g)(B_i) \subseteq V_i = \mathbb{C} B_i$. Thus we have in matrix form $$ [\phi(g)]_B = \begin{bmatrix} [\phi_1(g)]_{B_1} & 0 \\ 0 & [\phi_2(g)]_{B_2} \end{bmatrix} $$

But why is the matrix of this form? Can someone give a detailed explanation? (I know this is linear algebra). Thank you in advance.

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"But why" is not a complete question. Which step above are you having a problem with? –  Thomas Andrews Feb 7 '13 at 14:35
    
Yes, I am sorry for the lack of clarity. It is the last statement about the matrix form (edited). –  fran.aubry Feb 7 '13 at 14:41

1 Answer 1

up vote 2 down vote accepted

First I assume that you mean that $\phi_i(g) = \phi(g)\lvert_{V_i}$

Say that $B_1 = \{e_1, \dots, e_n\}$ and $B_2 = \{f_1, \dots, f_m\}$. Then, as you note, $B = B_1 \cup B_2$. With respect to this basis you want to write down the matrix for $\phi(g)$. The way you (by definition) do this is:

  • The first column in the matrix is the (vertical) vector you get by finding $\phi(g)e_1 = \phi_1(g)e_1$
  • The second column in the matrix is the (vertical) vector you get by finding $\phi(g)e_2 = \phi_1(g)e_2$
  • And so on
  • The last column in in the matrix is the (vertical) vector you get by finding $\phi(g)f_m = \phi_2(g)f_m$

Now you have have noted that $\phi(g)B_1 \subseteq \mathbb{C}B_1$ so that means that the vectors in the first $n$ columns only have something non-zero in the first $n$ entries (rows). That is $\phi(g)e_i \subseteq \mathbb{C}B_1$.

Likewise $\phi(g)B_2\subseteq \mathbb{C}B_2$, so the vectors in the last $m$ columns on the matrix will only have something non-zero in the last $m$ entries (rows).

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