# Find a basis such that the representation is diagonal

Here's an interesting question I'm working on for my algebra class.

Let $$G$$ be a finite group and let $$V$$ be a finite dimensional $$C G$$-module. Suppose that we have an isomorphism of $$CG$$-modules $$V \cong U_1 \oplus U_2 \oplus \cdots \oplus U_k,$$ where each of the $$U_i$$ is $$1$$-dimensional. Prove that it is possible to choose a basis of $$V$$ so that $$V$$ affords a representation $$\rho$$ with the property that for all $$g \in G, \rho(g)$$ is a diagonal matrix.

I am struggling to formulate a proof, but here are my current ideas:

Because we can "factor" $$V$$ into submodules that are all single dimensional, we choose a diagonal matrix with values of all the eigenvalues of the $$U_i$$. Now, this will be a representation because it will have the same eigenvalues as $$\rho$$ and thus are sort of "equivalent representations," but the one we are after is diagonal.

Is this concept correct?

• Feb 19 at 23:58

This is essentially by the definitions. The main difficulty is of course that there are just a ton of definitions. Let us review: there are 3 languages used in representation theory:

• We start with a $$G$$-space $$V$$, i.e. a (complex, I guess in your case, but generally over any field) vector space $$V$$, with a group action by $$G$$ on $$V$$ respecting the linear structure of $$V$$ (so $$g \bullet (\lambda v) = \lambda \cdot (g \bullet v)$$, where $$\cdot$$ denotes scalar multiplication of vectors by scalars in the vector space $$V$$ and $$\bullet$$ represents the group action by $$g\in G$$ on vectors $$v\in V$$. Equivalently, one can think of this as $$\color{red}{\text{"G acting on basis vectors \cal B:=\{v_1,\ldots, v_n\} of V, inside V"}}$$ (in the sense that $$g \bullet v_i\in V$$ for every $$g$$; and for every $$i$$, the action of $$G$$ on $$v_i$$ respects the group structure of $$G$$, like in the group action axioms; however this isn't exactly a group action on the set $$\cal B$$ because $$G$$ doesn't map $$\cal B$$ to itself, i.e. $$g \bullet v_i$$ needn't be in $$\cal B$$, just in $$V$$), extended to a $$G$$-action on all of $$V$$ by linearity.

• The $$G$$-action on $$V$$ can be encoded by a group homomorphism $$\rho: G \to GL(V)$$, i.e. by $$\rho(g)v := g \bullet v$$. (That $$\rho(g)$$ is an endomorphism of $$V$$, i.e. a linear map/transformation $$V\to V$$ is by the $$G$$-action "respecting linearity" as I mentioned above; and that $$\rho$$ is a group homomorphism is by the group action axioms).

• We can also extend the $$G$$-action by linearity to a $$\mathbb C[G]$$-action on $$V$$, i.e. a $$\mathbb C[G]$$-module structure on $$V$$. Although the difference between a $$G$$-space $$V$$ and a $$\mathbb C[G]$$-module $$V$$ seems artificial/largely a matter of syntax ("we're just writing a bunch of scalars that don't do anything new"), the change in perspective allows us to think of representations in terms of module theory, i.e. turn the problem of group representations into a more complicated version of linear algebra (which is a good thing, since people understand linear algebra quite well).

Given anything in one of the 3 languages, we can translate to the other 2: for example, given a representation $$\rho : G \to GL(V)$$, or sometimes just stated "an $$n$$-dimensional representation $$\rho$$", we have the $$\mathbb C$$-vectorspace $$V$$ (or if you're just given the dimension, $$V = \mathbb C^n$$), and then $$\rho$$ gives us a $$G$$-action on this $$V \simeq \mathbb C^n$$, making it a $$G$$-space.

Ok, let's understand the $$\mathbb C[G]$$-modules $$U_1,\ldots, U_k$$. These are 1-dimensional, meaning 1-dimensional as $$\mathbb C$$-vectorspaces, i.e. for any $$u_i \neq 0_{U_i}$$ we have $$U_i = \mathbb C \cdot u_i := \{\lambda u_i: \lambda\in \mathbb C\}$$ (so $$u_i$$ is a basis vector of $$U_i$$). By my first bullet point, one can understand the $$\mathbb C[G]$$-module of $$U_i$$ as just $$\color{red}{\text{"G acting on u_i, inside U_i"}}$$ (in the sense explained above). If you think of $$U_i$$ as the complex plane with an origin (this is the picture of a 1-dimensional $$\mathbb C$$-vectorspace), and $$u_i$$ as a vector of $$U_i$$ (not the origin), $$G$$ is just a collection of ways you can move $$u_i$$ in $$U_i$$ (group axioms mean you can compose movements, and that all movements can be exactly undone).

Next, that $$V$$ is isomorphic to $$U_1 \oplus \ldots \oplus U_k$$ as $$\mathbb C[G]$$-modules in particular means that they are isomorphic as $$\mathbb C$$-vectorspaces. Let $$v_1,\ldots, v_k\in V$$ be the isomorphic copies of $$u_1,\ldots, u_k$$ (which recall are the basis vectors of $$U_1,\ldots, U_k$$). The definition of direct sum (of $$\mathbb C$$-VS) tells us that $$v_1,\ldots, v_k$$ form a basis of the $$\mathbb C$$-VS $$V$$. (If you want notation, let $$\Phi$$ be the $$\mathbb C[G]$$-isomorphism from the direct sum to $$V$$. Then $$v_i = \Phi(u_i)$$, or rather $$v_i = \Phi((0,\ldots, 0, u_i,0,\ldots, 0))$$ since $$\Phi$$ is defined on $$U_1 \oplus \ldots \oplus U_k$$ --- but from now on we will abuse notation in this manner. The 1st sentence of this paragraph used that $$\Phi$$ is $$\mathbb C[G]$$-linear $$\implies \Phi$$ is $$\mathbb C$$-linear.)

Let's add back the $$G$$-structure (i.e. that $$\Phi$$ respects action by $$G$$). In particular, we have $$g \bullet u_i = \lambda_{g,i} \cdot u_i \implies g \bullet v_i = g \bullet \Phi(u_i) = \Phi(g \bullet u_i) = \Phi(\lambda_{g,i} \cdot u_i)= \lambda_{g,i}\cdot \Phi(u_i) =\lambda_{g,i} \cdot v_i$$, i.e. all $$G$$ send $$v_i$$ to something in the $$\mathbb C$$-subspace of $$V$$ spanned by $$v_i$$. This should be no surprise (although the formal calculation is long), since $$G$$ preserved the space $$U_i$$ (i.e. mapped $$U_i$$ to itself), so naturally because $$\Phi$$ preserves all that structure, $$G$$ should preserve the space $$\text{span}_{\mathbb C}\{v_i\}= \Phi(U_i)\subseteq V$$ (the image of $$U_i$$ in $$V$$ under $$\Phi$$, or to be pedantic, the image of $$\{0\}\times \ldots\times \{0\}\times U_i \times \{0\}\times\ldots \times \{0\} \subseteq U_1 \oplus \ldots \oplus U_k$$). "$$G$$ preserving a given subspace of $$V$$" is also stated as "the given subspace of $$V$$ is $$G$$-invariant".

So indeed for this basis $$\cal B:=\{v_1,\ldots, v_k\}$$ of $$V$$, we have found that the $$G$$-action, encoded by $$\rho:G \to GL(V)$$ maps $$\rho(g)v_i = \lambda_g \cdot v_i$$. Or in other words, writing vectors in $$V$$ w.r.t. the basis $$\cal B$$ we have $$\rho(g)(0,\ldots, 0,1,0,\ldots, 0) = (0,\ldots, 0, \lambda_{g,i}, 0,\ldots, 0)$$. So writing $$\rho(g)$$ as a matrix, it is indeed diagonal with entries $$\lambda_{g,i}$$ ranging from $$1\leq i\leq k$$ on the diagonal.

Summary: essentially all the definitions in this problem boil down to saying that $$V$$ can be decomposed as $$k$$ 1-dimensional subspaces that are all $$G$$-invariant, so in the basis of $$V$$ formed by picking a non-zero vector in each of those $$k$$ 1-dimensional subspaces ("lines"), any $$g\in G$$ can only act by scaling (since those are the only linear transformations in each of the 1-dimensional subspaces). And this is the definition of a diagonal matrix w.r.t. that basis.

Reflections: although what I wrote was very long winded, it was nothing more than just going through the horde of definitions carefully. The definitions involved were:

• 3 languages of representation theory and all definitions therein ($$G$$-actions, $$G$$-spaces, respecting ... structure, basis of vector space, group homomorphism, endomorphisms of a vector space, modules, the group ring $$\mathbb C[G]$$);
• 1-dimensional $$\mathbb C[G]$$-modules
• direct sums of modules/vectorspaces, $$\mathbb C[G]$$-modules isomorphisms, $$G$$-invariant subspaces of $$V$$
• subspaces/span in vector spaces, image of maps,
• translation between abstract vector spaces and concrete representation of vector spaces and transformations (in the form of lists of numbers/matrices of numbers) w.r.t. a given basis, diagonal matrices.

It is no wonder that students are confused after learning representation theory for the first time!

• Thanks for such a thorough answer, Dr. Rui. Feb 20 at 17:27