We know the quantum circuit representation of Fourier operator. Now, I want the quantum circuit for the following unitary matrix using single and two qubit gates: $$U = \begin{pmatrix} F_3 & 0 \\ 0 & 1 \end{pmatrix}, $$ where $$F_3 = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega \\ \end{bmatrix}$$ is the $3\times 3$ Fourier matrix and $w = e^{2\pi i / 3}$. The operator $U$ acts of $2$-qubit states and keeps $|11 \rangle$ unchanged.
In general, what is the circuit representation of the following operator
$$U =
\begin{bmatrix}
F_m & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1(2^n - m)\text{- posiiton}
\end{bmatrix},
$$
where
$$F_m = \frac{1}{\sqrt{m}}
\begin{bmatrix}
1 & 1 & \cdots & 1 \\
1 & \omega & \cdots & \omega^{m-1} \\
\vdots & \vdots & \ddots & \vdots \\
1 & \omega^{m-1} & \cdots & \omega^{(m-1)(m-1)}
\end{bmatrix}?, \quad \omega = e^{2 \pi i / m}$$ Here, $\omega$ represents a primitive $m$-th root of unity.
