the linear span of all matrix coefficients is $C(G,\mathbb{C})$ where $G$ is a finite group

Theorem. Let $\{(R_{\alpha},V_{\alpha})\}$ be a complete set of inequivalent irreducible finite dimensional representations of a finite group $G$. Let $V_{R_{\alpha}}$ be the subspace generated by all matrix coefficients of $R_{\alpha}$ Then the spaces $V_{R_{\alpha}}$ are mutually orthogonal. Moreover, they generate the space $C(G,\mathbb{C})$

Question: In the following proof, is it necessary to show that the sum $\bigoplus_{\alpha}V_{R_{\alpha}}$ is nonzero? Thanks.

Proof: We show that $\bigoplus_{\alpha}V_{R_{\alpha}}=C(G,\mathbb{C})$. Suppose the sum is a proper subspace of $C(G,\mathbb{C})$. If the sum is zero, then $V_{R_{\alpha}}=0$ for all $\alpha$. Then all matrix coefficients of $R_{\alpha}$ is zero i.e. $R_{\alpha}(x)\in GL_{n}(\mathbb{C})$ is a zero matrix for all $x\in G$, contradiction. Hence the sum is nonzero. Now we show that for any $\alpha$, the space $V_{R_{\alpha}}$ is invariant under the right regular representation $r$ :

For any $g\in G$ and any matrix coefficient $f_{u,v}\in V_{R_{\alpha}}$, we have $$\big(r(g)f_{u,v}\big)(x)=f_{u,v}(xg)=\langle R_{\alpha}(xg)u,v\rangle=\langle R_{\alpha}(x)R(g)u,v\rangle=f_{R(g)u,v}$$ Hence the set of all matrix coefficients of $R_{\alpha}$ is invariant under $r$. By the linearity of $r(g)$, the space $V_{R_{\alpha}}$ is also invariant. Therefore the sum is invariant under $r$. We know that the orthogonal complement $W$ of the sum is invariant under $r$. We may choose an invariant subspace $U$ of $W$ such that $r$ is irreducible on $U$.

Let $\{u_1,...,u_m\}$ be an orthonormal basis of $U$. Then each function $x\mapsto \langle r(x)u_j,u_i\rangle$ is orthogonal to $U$ by construction i.e. $$0 =\sum_{x\in G}\langle r(x)u_j, u_i\rangle\overline{f(x)} \qquad \text{for all f\in U}$$ Consider the linear functional $L:U\to\mathbb{C}$ defined by $L(f)=f(1)$. Applying the Riesz Representation Theorem, there is an element $f_0\in U$ such that $L(f)=\langle f, f_0\rangle$.On the other hand, for any $f\in U$ $$f(x)=f(1.x)=(r(x)f)(1)=L(r(x)f)=\langle r(x)f, f_0\rangle$$ Then we have $$0=\sum_{x\in G}\langle r(x)u_j,u_i\rangle\overline{\langle r(x)f, f_0\rangle}=\frac{|G|\langle u_j, f\rangle\overline{\langle u_i, f_o\rangle}}{\dim U}$$ for all $i$ and $j$. Since we can take $f=u_j=u_1$ and since $i$ is arbitrary, this equation forces $f_0=0$ and gives a contradiction.

• Well, it is nonzero because there is the trivial representation (which sends every element of $G$ to $1 \in \mathbb{C})$). – Geoff Robinson Jan 24 '16 at 11:43
• Is it logically necessary to indicate this in the proof? Thanks – Ergin Suer Jan 24 '16 at 11:46
• It depends how pedantic you want to be. Strictly speaking, yes, you need to know that the vector space you want to prove has certain properties is actually a vector space in the first place, so is non-empty in particular. – Geoff Robinson Jan 24 '16 at 11:50
• I agree that this comes down to how pedantic you want to be. I think a maximal pedant would say that the proof is already complete if you delete all lines between "If the sum is zero...Hence the sum is nonzero.". It is clear that the $0$-function is in the space, so the sum is a vector space; the question is whether it is the vector space whose sole element is $0$. If you work with the assumption that it is $0$-vector space, then every step of the proof is correct until, at the end, you deduce that $\oplus V_{R_a} = \mathbb{C}^G$ (and, in particular, is not the $0$-vector space.) – David E Speyer Jan 24 '16 at 15:24