 Solution

Write the lower block of $T(a)$ as $B=\left(\begin{array}{cc}0& b\\ c& 0\end{array}\right)$. ThenA representation of $C_4$ must satisfy its defining relation $T(a{)}^4=I_3$, so $(bc{)}^2=1$. For real $b,c$, this means $bc=1$ or $bc=-1$; it also ensures invertibility.

If $bc=-1$, then $T(a{)}^2\ne I_3$ while $T(a{)}^4=I_3$. The image of the generator has order four, so its four powers are distinct and the representation is faithful. If $bc=1$, then $T(a{)}^2=I_3$ but $T(a)\ne I_3$, since the lower block has zero diagonal. The image has order two and the kernel is $\{I,a^2\}$. HenceFor other values there is no representation of $C_4$, rather than an additional unfaithful case; in particular $bc=0$ would make the assigned matrix singular. These conclusions are the faithfulness criterion for a block representation of the cyclic group of order four. There is no requirement that the proposed representation be orthogonal, so no separate conditions $|b|=|c|=1$ should be imposed.