# One question on block-circulant matrices

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 $$n \times n$$.

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 $$1$$.
• @PadraigÓCatháin I don't think so. I don't use the multiplication of blocks. The formula $K=U^*DU$ where $D$ is block-diagonal is always valid. And it is known that $\det D$ is the product of the determinants of the diagonal blocks. Nov 3, 2022 at 12:35