Solution
= Solution
Write
$$
u(x)=\begin{pmatrix}1&x\\0&1\end{pmatrix}.
$$
Since $u(x)u(y)=u(x+y)$,
$$
\begin{aligned}
\chi(v\mathbin\cdot)(u(x+y))
&=\chi(u(vx+vy))\\
&=\chi(u(vx))\chi(u(vy))\\
&=\chi(v\mathbin\cdot)(u(x))\chi(v\mathbin\cdot)(u(y)).
\end{aligned}
$$
\b[Thus $\chi(v\mathbin\cdot)$ is a one-dimensional representation of $U$; for $v=0$ it is the trivial representation.]