# Inverse of circulant matrices

The following is an n x n circulant matrix where h is real but not equal to 1:

$$A= \begin{bmatrix} 2&-h&0&0&...&...&0&-h\\ -h&2&-h&0&...&...&0&0\\ 0&-h&2&-h&...&...&...&...\\ 0&0&-h&2&...&...&...&...\\ ...&...&...&...&...&...&0&0\\ ...&...&...&...&...&...&-h&0\\ 0&0&...&...&0&-h&2&-h\\ -h&0&...&...&0&0&-h&2 \end{bmatrix}$$

So my question is how can I find an explicit formula for the inverse of this matrix?

So for $M = QDQ^T$, I found that $D = \mathrm{diag}[2(1-h*\cos(2\pi l/N))]$ from $0$ to $N-1$ somehow by reading some papers on unitary Van der Monde matrices and shift matrices. However, I don't fully understand how this works.

Can someone please show me how to derive an explicit formula for the inverse of this matrix and explain why we are able to do so?

Appreciate any help in advance.

As you already know (see also Wiki:Circulant matrices), you can diagonalize your matrix $A=QDQ^T$, with $Q$ being the discrete Fourier Transform , a special kind of Vandermonde matrix.
To get the inverse of $A$, all you got to do is to invert all diagonal entries of $D$, i.e. $d_{ij} \mapsto \frac1{d_{ij}}$.