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I am having the following block matrix

$$\left[\begin{array}{cccc} \mathbf{A+B} & \mathbf{B} & \cdots & \mathbf{B} \\ \mathbf{B} & \mathbf{A+B} & \cdots & \mathbf{B} \\ \vdots & \vdots & \ddots & \vdots\\ \mathbf{B} & \mathbf{B} & \cdots & \mathbf{A+B} \end{array}\right]$$

where $\mathbf{A}$ and $\mathbf{B}$ are full rank, symmetric square matrices. There are $n$ blocks in each direction. I want to obtain the determinant of the block matrix.

I play with some examples and the determinants seems to be

$$\det(\mathbf{A})^{n-1}\det(\mathbf{A}+n\mathbf{B})$$

May I ask whether this is correct or not, and is there any proof?

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2 Answers 2

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the above-mentioned link uses to end up the proof a topological argument of density. Here is a simple purely algebraic proof, based on operations on rows and on columns:

Substracting the last block-row from the $n-1$ first block-rows yields: $$\begin{bmatrix} A&0&0&\dots&0&-A\\ 0&A&0&\dots&0&-A\\ 0&0&A&\dots&0&-A\\ \vdots&&&&&\vdots\\ 0& 0&0&\dots&A&-A \\ B&B&B&\dots&B&A+B \end{bmatrix}$$ Now add each of the $n-1$ first columns to the last one, to get: $$\begin{bmatrix} A&0&0&\dots&0&0\\ 0&A&0&\dots&0&0\\ 0&0&A&\dots&0&0\\ \vdots&&&&&\vdots\\ 0& 0&0&\dots&A&0\\ B&B&B&\dots&B&A+nB \end{bmatrix}$$ We have a lower block-triangular matrix. Its determinant is the product of the determinants of the diagonal blocks: $$\lvert A\rvert^{n-1}\cdot\lvert A+nB\rvert.$$

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Your matrix is equal to $E\otimes B+I\otimes A$ (where $E$ is the all-one matrix), which is similar to $nE_{11}\otimes B+I\otimes A=\operatorname{diag}(nB+A,\,A,\,\ldots,\,A)$ (where $E_{11}$ is the matrix whose only nonzero entry is a $1$ at the $(1,1)$-th position). Hence the result.

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