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Let $N$ be the normal closure of the upper unipotents $U(t)=A(t)$. Conjugation by $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ givesThus $N$ also contains every lower unipotent $L(s)=\left(\begin{array}{cc}1& 0\\ s& 1\end{array}\right)$. We prove explicitly that these elementary unipotent generators of SL2R generate the full group.

For $a\ne 0$, multiplication givesSo all determinant-one diagonal matrices lie in $N$. If $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ has $a\ne 0$ and $ad-bc=1$, thenbecause its lower-right entry is $(1+bc)/a=d$. If $a=0$, then $c\ne 0$ and left multiplication by $U(1)$ makes the upper-left entry equal to $c$. The resulting matrix lies in $N$ by the preceding factorization, and $U(1{)}^{-1}\in N$ restores $g$. Therefore

The diagonal element $\mathrm{diag}(m,m^{-1})$ conjugates $U(t)$ to $U(m^{2}t)=U(t{)}^{m^2}$. In a finite-dimensional unitary representation, conjugacy therefore makes its eigenvalue multiset invariant under every square power. Square-power rigidity of a finite nonzero spectrum makes the image of every upper unipotent the identity. Its kernel is a normal subgroup; elementary unipotent generators of SL2R then force the entire group representation to be trivial. This algebraic argument does not require continuity.