Inverse of Hermitian matrix which is constant on the off-diagonal For $N\in\mathbb N$, I am interested in the inverse of the matrix
$$M=\left(\begin{matrix}N&a&\dots&\dots&a\\
\bar a&N & a&\dots&a\\ 
\vdots &\ddots & \ddots&\ddots&\vdots\\
\bar a& \dots & \bar a&N&a\\
\bar a &\dots&\dots&\bar a &N  \end{matrix}\right) \in \mathbb C^{N\times N}$$ where $a\in\mathbb C$ with $\lvert a\rvert<1$. That is, $M$ is a strictly diagonal dominant matrix with $a$ above the diagonal, $\bar a$ below the diagonal and $N$ on the diagonal. More specifically, I am interested in $$\langle e,M^{-1}e\rangle$$ where $e=(1,\dots,1)^T\in\mathbb R^N$. 
If $a$ is real, everything is rather simple: We can write $$M^{-1}=[(N-a)1+a\, ee^T]^{-1}=\frac{1}{N-a}1-\frac{1}{N-a}\frac{a\, ee^T}{N-a+a\,e^Te}=\frac{1}{N-a}\left[1-\frac{a}{(1+a)N-a}ee^T\right],$$ for example by the Sherman-Morrison formula. Anything that can be done for complex $a$?
 A: (This is too long for a comment.)
When $N=3$, the expression of $M^{-1}$ is already rather messy. I'm not sure if there is any simple formula for $M^{-1}$ with a general $N$. Note that the complex case is very different from the real case. In the real case, $M$ is just a rank-one update of a scalar matrix, but in the complex case, $M$ is almost never such a rank-one update, because it almost always has $N$ distinct eigenvalues.
Anyway, your $M$ is a Toeplitz matrix. As there is a huge volume of research on Toeplitz matrices, you may search the internet or any research database for relevant results.
In general, the inverse of a Toeplitz matrix is not Toeplitz, but in your case, it is. Here is the reason. First, let $b=\bar a$ and
$$
T=\pmatrix{
b&a&\dots&\dots&a\\
b&\ddots&\ddots&\ddots&\vdots\\ 
\vdots&\ddots&\ddots&\ddots&\vdots\\
\vdots&\ddots&\ddots&\ddots&a\\
b&\dots&\dots&b&b},
\quad C=\pmatrix{
0&&&&-\frac{a}{b}\\
-1&\ddots\\ 
&\ddots&\ddots\\
&&\ddots&\ddots\\
&&&-1&0}.
$$
It is straightforward to verify that $T^{-1}=\frac1{b-a}(I+C)$. Now, by Cayley-Hamilton, $M^{-1}$ is a polynomial in $M=(N-b)I+T$. Hence it is also a polynomial in $T$, a polynomial in $T^{-1}$ and a polynomial in $C$. It is easy to verify that every power of $C$ is a Toeplitz matrix. Therefore $M^{-1}$ must be Toeplitz too.
