Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
"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
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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.