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 $n \times n$.


1 Answer 1


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 $1$.

  • $\begingroup$ Block-circulant can be used in two senses: that the blocks are circulant, or that the template in which the blocks are placed is circulant. Your formula requires both senses! In general, there are formulas for determinants of a block matrix in terms of polynomial expressions in the blocks provided that the blocks commute. $\endgroup$ Nov 3, 2022 at 10:45
  • 1
    $\begingroup$ @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. $\endgroup$ Nov 3, 2022 at 12:35

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