Consider the regular representation $V$ of the cyclic group of order 3, $G=\{ e,a,a^2\}$, with the standard basis $\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$. Consider then another basis given by $$\tag{1} (\textbf{v}_1\textbf{u}_1,\textbf{u}_2):=\left(\frac{1}{\sqrt{3}}(\textbf{e}_1+\textbf{e}_2+\textbf{e}_3),\frac{1}{\sqrt{2}}(\textbf{e}_1-\textbf{e}_2), \frac{1}{\sqrt{6}}(\textbf{e}_1+\textbf{e}_2-2\textbf{e}_3)\right), $$ where $\{ \mathbf{u}_1,\mathbf{u}_2 \}$ defines a basis for a submodule $U$ of $V$. Now, $G$ clearly acts trivially on the $G$-module spanned by $\mathbf{v}_1$, so the representation of that submodule is the $1\times 1$ identity matrix. Is the same true for the quotient module, $V/U$?
Edit: the representation of the group with the above basis is $$\tag{2} \sigma(e)= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad \sigma(a)= \begin{bmatrix} 1 & 0 & 0 \\ 0 & -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}, \qquad \sigma\left(a^2\right)= \begin{bmatrix} 1 & 0 & 0 \\ 0 & -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ 0 & -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}. $$ According to Algebra by P.M. Cohn, if $V'$ is a submodule of a module $V$ and the basis $\{ \mathbf{v}_1,...,\mathbf{v}_f \}$ is written such that the last members form a basis for $V'$, then the representation for $V$ takes the form: $$ \rho(x)=\begin{pmatrix}\rho''(x)&\theta(x)\\ 0&\rho'(x)\end{pmatrix} $$ where $\rho'(x)$ is a representation for $V'$ and $\rho''$ a representation for the quotient module $V/V'$. With this in mind, I would assume that the upper $1\times 1$ matrix of eq. $(2)$ is a representation for $V/U$. What am I misunderstanding?