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Put $U=\varphi (A(t))$. Applying the group homomorphism $\varphi$ to the identity just proved shows that $U$ is conjugate to $U^{m^2}$ for every positive integer $m$. Thus the multiset of its eigenvalues is unchanged by taking their $m^2$th powers.

Take $m=2$. The fourth-power operation permutes this finite multiset. For each eigenvalue $z$, iterating its permutation cycle gives $z^{4^r}=z$ for some positive integer $r$. Since $U$ is unitary, $z\ne 0$, so $z^{4^r-1}=1$: every eigenvalue is a root of unity. Choose a positive integer $d$ divisible by all their orders. Every eigenvalue of $U^{d^2}$ is then one. Because its eigenvalue multiset is also the original one, we obtainThis is square-power rigidity of a finite nonzero spectrum. Finiteness of the dimension is essential to the permutation-cycle argument.

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.