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Consider the $k \times k$ block matrix:

$ C = \left(\begin{array}{ccccc} A & B & B & \cdots & B \\ B & A & B &\cdots & B \\ \vdots & \vdots & \ddots & \cdots & \vdots \\ B & B & B & \cdots & A \end{array}\right) = I_k \otimes (A - B) + \mathbb{1}_k \otimes B$

where $A$ and $B$ are size $n \times n$ and $\mathbb{1}$ is the matrix of all ones. It would seem that the formula for the determinant of $C$ is simply:

$\det(C) = \det(A-B)^{k-1} \det(A+(k-1) B)$

Can anyone explain why this seems to be true or offer a proof or direct me to a proof?

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Diagonalize $1_k$. –  Federico Poloni Aug 26 at 8:58

1 Answer 1

up vote 8 down vote accepted

We can just manipulate $C$ in the usual way by row operations: Subtract the last "row" from all the other "rows" (this is really several traditional row operations done at once). This produces $$ \begin{pmatrix} A- B &0& 0 & \ldots & 0 &B-A \\ 0 & A-B &0 &\ldots & 0 & B-A\\ && \ldots &&&\\ B & B & B & \ldots & B & A \end{pmatrix} . $$ Assume for the moment that $A-B$ is invertible. Subtract $B(A-B)^{-1}$ times all the other rows from the last row; we multiply from the left so that we indeed obtain linear combinations of the rows. This gives an upper triangular matrix with diagonal entries $A-B$ ($k-1$ times) and $A+(k-1)B$. We now read off the asserted formula.

The invertible matrices are dense, so I obtain the general case by approximation.

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Fantastic, Thanks. –  amcalde Aug 25 at 19:22

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