Brian Bi

## Section 3.3. Representations of direct sums of matrix algebras

Problem 3.3.3

1. We will prove the following:

Lemma: If $$V$$ is a representation of $$A$$, where $$A = \bigoplus_{i=1}^n A_n$$, then $$V$$ can be written as the direct sum of representations $$V = \bigoplus_{i=1}^n 1_i V$$.

Proof: Suppose $$v \in V$$ is given. Since $$1 = \sum_{i=1}^n 1_i$$, it follows that $$v = \sum_{i=1}^n 1_i v$$. Also, suppose $$v = \sum_{i=1}^n v_i$$ where $$v_i \in 1_i V$$ for all $$i$$. Then, let $$j \in \{1, 2, \ldots, n\}$$; left-multiplying by $$1_j$$ yields $$1_j v = \sum_{i=1}^n 1_j v_i = \sum_{i=1}^n 1_j 1_i v_i = v_i$$. Therefore, $$V$$ is the direct sum of the $$1_i V$$'s as vector spaces. But each $$1_i V$$ is also a subrepresentation of the full $$A$$ (where the $$A_j$$ components act as zero for all $$j \neq i$$), which proves the desired result.

Corollary: $$V$$ is irreducible if and only if there is exactly one $$i$$ for which $$1_i V$$ is nonzero, and that $$1_i V$$ is an irrep of $$A_i$$.

Proof of corollary: Write $$V = \bigoplus_{i=1}^n 1_i V$$. If there is a unique $$i$$ with $$1_i V$$ nonzero, then $$V = 1_i V$$. Since $$A$$ acts on $$1_i V$$ exactly as its $$A_i$$ component does, it follows that $$V$$ is an irrep of $$A$$ iff $$1_i V$$ is an irrep of $$A_i$$. If, on the other hand, there are multiple $$i$$'s with $$1_i V$$ nonzero, then $$V$$ is not irreducible, since a nonzero proper subrepresentation is constructed by removing one $$1_i V$$ component from the direct sum.

Therefore, we can find all irreps of $$A$$ by considering each $$i$$ in turn as the $$i$$ for which $$1_i V \neq \{0\}$$. For each such pair $$(i, V)$$ where $$V$$ is an irrep of $$A_i$$, $$V$$ is also an irrep of $$A$$ with $$(a_1, \ldots, a_n)v = a_i v$$; and all irreps of $$A$$ are of this form.

2. Following the Hint, for every $$(i, j) \in \{1, 2, \ldots, d\}^2$$, let $$E_{ij} \in \mathrm{Mat}_d(k)$$ be the matrix with 1 in the $$i$$th row of the $$j$$th column and 0's everywhere else. Let $$V$$ be a finite-dimensional representation of $$\mathrm{Mat}_d(k)$$, and let $$v \in V$$. Now since $$E_{11} + \ldots + E_{dd}$$ is the identity, it follows that $$v = E_{11}v + \ldots + E_{dd}v$$. So each $$v \in V$$ can be written as $$v = \sum_{i=1}^d v_i$$ where $$v_i \in E_{ii}V$$ for every $$i$$. Furthermore, this decomposition is unique since left-multiplying by $$E_{jj}$$ yields $$E_{jj} v = v_j$$ for each $$j$$. We conclude that $$V = \bigoplus_{i=1}^d E_{ii}V \label{eqn:matdirectsum}$$

Also, for every $$v \in E_{11}V$$, and for every $$i \in \{1, 2, \ldots, d\}$$, we have $$E_{ii}E_{i1}v = E_{i1}v$$, so $$E_{i1}v \in E_{ii}V$$. Let $$\Phi_i: E_{11}V \to E_{ii}V$$ be defined by $$\Phi_i(v) = E_{i1}v$$. Observe that if $$w \in E_{ii}V$$, then the equation $$w = E_{i1}v$$ has exactly one solution: left-multiplying by $$E_{1i}$$ yields $$v = E_{1i}w$$, and indeed $$E_{i1}(E_{1i}w) = E_{ii}w = w$$. Therefore $$\Phi_i$$ is one-to-one and onto.

Fix $$v \in E_{11}V \setminus \{0\}$$, and let $$S(v) = \langle E_{11}v, E_{21}v, \ldots, E_{d1}v\rangle$$. Since each $$\Phi_i$$ is one-to-one, it follows that none of the $$E_{i1}v$$'s vanish, and since each lies in $$E_{ii}V$$, it follows from $$(\ref{eqn:matdirectsum})$$ that they are all linearly independent. So $$S(v)$$ has dimension $$d$$, and can be put into isomorphism with $$k^d$$ (with standard basis $$\{e_1, \ldots, e_d\}$$), with the mapping given by $$E_{i1}v \mapsto e_i$$. If $$a \in \mathrm{Mat}_d(k)$$, write $$a = \sum_{j,k} c_{jk} E_{jk}$$; then for each $$i$$, $$a E_{i1} v = \sum_{j,k} c_{jk} E_{jk} E_{i1}v = \sum_j c_{ji} E_{j1}v \in S(v)$$, so $$S(v)$$ is a subrepresentation of $$V$$; also, since $$a e_i = \sum_{j,k} c_{jk} E_{jk} e_i = \sum_j c_{ji} e_j$$, it follows that $$S(v)$$ and $$k^d$$ are in fact isomorphic as representations. So $$V$$ is irreducible iff $$V$$ is isomorphic to $$k^d$$ as a representation of $$\mathrm{Mat}_d(k)$$. (Since $$\mathrm{Mat}_d(k)$$ is finite-dimensional, it cannot, of course, have any infinite-dimensional irreps either.)

Fix a basis $$\{v_1, v_2, \ldots, v_k\}$$ of $$E_{11}V$$. Suppose $$v \in V$$ is given. Using $$(\ref{eqn:matdirectsum})$$ and the fact that each $$\Phi_i$$ is an isomorphism, we see that there is a unique set of coefficients $$c_{ij}$$ such that $$v = \sum_{j=1}^k \sum_{i=1}^d c_{ij} E_{i1} v_j$$. Therefore $$V = S(v_1) \oplus S(v_2) \oplus \ldots \oplus S(v_k)$$ as vector spaces. And since each $$S(v_j)$$ is a subrepresentation of $$V$$, we conclude that $$V = S(v_1) \oplus \ldots \oplus S(v_k)$$ as representations.

3. Let $$A = \bigoplus_{i=1}^r A_i$$ where $$A_i = \mathrm{Mat}_{d_i}(k)$$. By part (a), every irrep of $$A$$ is obtained as an irrep of one of the $$A_i$$'s. By part (b), the only irrep of $$A_i$$ is $$k^i$$. Therefore all irreps of $$A$$ are given by $$V_1 = k^{d_1}, V_2 = k^{d_2}, \ldots, V_r = k^{d_r}$$. If $$V$$ is a general finite-dimensional representation of $$A$$, then by the Lemma proven in part (a), we can write $$V = \oplus_{i=1}^r 1_i V$$. Each $$1_i V$$ is a representation of $$A_i$$ so by part (b), it decomposes as a direct sum of $$k^{d_i}$$'s, and we are done.