Operator Norm of a Matrix composed of Standard Basis and Fourier Basis Let $\mathbf{A}_n$ be an $n\times 2n$ matrix (where $n=2^k$) composed of Fourier basis and standard basis; that is, 
$$\mathbf{A}_n = \begin{bmatrix}\mathbf{I}_n & \mathbf{F}_n\end{bmatrix}$$
where $\mathbf{I}_n$ is the identity matrix and $\mathbf{F}_n$ is the DFT matrix.
Generally, a DFT matrix $\mathbf{F}_n$ is defined as 
$$
\mathbf{F}_n = \frac{1}{\sqrt{n}}
\begin{bmatrix}
\omega^0 & \omega^0 & \omega^0 & \cdots & \omega^0 \\
\omega^0 & \omega^1 & \omega^2 & \cdots & \omega^{n-1} \\
\omega^0 & \omega^2 & \omega^4 & \cdots & \omega^{2(n-1)} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\omega^0& \omega^{n-1}& \omega^{2(n-1)}& \cdots & \omega^{(n-1)(n-1)} 
\end{bmatrix}
$$
where $\omega = \exp(-2\pi i / n)$ is the primitive $n^\mathrm{th}$ root of unity.
For example, if $n=2^2$, $\mathbf{A}_4$ is the concatenation of $\mathbf{F}_4$ and $\mathbf{I}_4$:
$$
\mathbf{A}_4=
\begin{bmatrix}
1 & 0 & 0 & 0 & 1/2 & 1/2 & 1/2 & 1/2 \\
0 & 1 & 0 & 0 & 1/2 & -i/2 & -1/2 & i/2 \\
0 & 0 & 1 & 0 & 1/2 & -1/2 & 1/2 & -1/2 \\
0 & 0 & 0 & 1 & 1/2 & i/2 & -1/2 & -i/2
\end{bmatrix}
$$
One can easily obtain $\mathbf{F}_4$ by plugging in $\omega = \exp(-2\pi i / 4) = -i$.

I'm interested in computing the operator norm of $\mathbf{A}$ defined as:
$$||\mathbf{A}||_{op} = \max_{||\mathbf{x}||_2 = 1} ||\mathbf{A}\mathbf{x}||_2$$
where $||\mathbf{x}||_2$ is the Euclidean norm of the vector $\mathbf{x}$.
Do we have a closed form of $||\mathbf{A}_n||_{op}$ for any $n$?

The singular values of $(I,F)$ are the square roots of the eigenvalues of $(I,F)(I,F)^\ast=I+FF^\ast$. Since the DFT matrix $F$ is unitary, $I+FF^\ast=2I$. Consequently, all singular values and in turn the operator $2$-norm of $(I,F)$ are equal to $\sqrt{2}$, regardless of the size $n$.
