Circulant matrices are very useful in digital image processing.
I found the general formula for determinant of circulant matrix.
But I think it is not suitable for block-circulant matrices.


For example, consider the formula for $\det(K)$,


where $$K = \left(\begin{array}{cccc} A & B & C & D \\
 D & A & B & C \\
 C & D & A & B \\
 B & C & D & A 
\end{array}\right) $$


and $A$ , $B$ , $C$ and $D$ are size $n \times n$.



One question on block-circulant matrices
The formula for the specific case is
$$\det K=\det(A+B+C+D)\det(A-B+C-D)\det(A+iB-C-iD)\det(A-iB-C+iD).$$
More generally, for a block-circulant matrix with $n$ square blocks $A\_0,\ldots,A\_{n-1}$, the formula is
$$\det K=\prod\_{\omega^n=1}\det(A\_0+\omega A\_1+\cdots+\omega^{n-1}A\_{n-1}).$$
To see this, observe that $K$ is block-diagonalisable,
$$K=U^\*{\rm diag}(A\_0+\alpha A\_1+\cdots+\alpha^{n-1}A\_{n-1},A\_0+\alpha^2 A\_1+\cdots+\alpha^{2(n-1)}A\_{n-1},\ldots)U$$
where $\alpha=\exp\frac{2i\pi}n$ and
$$U=\frac1{\sqrt n}((\alpha^{(i-1)(j-1)}I\_d))\_{1\le i,j\le n}.$$
Hereabove, the blocks $A\_j$ are $d\times d$. This shows the formula, up to the factor $|\det U|^2$, which is easily seen to be equal to $1$.


