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I was messing around with some matrices and found the following result.

Let $A_n$ be the $(2n) \times (2n)$ matrix consisting of elements $$a_{ij} = \begin{cases} 1 & \text{if } (i,j) \leq (n,n) \text{ and } i \neq j \\ 1 & \text{if } (i,j) > (n,n) \text{ and } i \neq j \\ 0 & \text{otherwise}. \end{cases} $$ Then, the determinant of $A_n$ is given by $$\text{det}(A_n) = (n-1)^2.$$

Example: $$A_2 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}, A_3 = \begin{pmatrix} 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \end{pmatrix},$$ with det$(A_2)$ and det$(A_3)$ being $1$ and $4$ respectively. I was wondering if anybody could prove this statement for me.

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marked as duplicate by user1551 linear-algebra Jan 25 '17 at 22:33

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  • $\begingroup$ I guess it's better to use matrix with zero-angle. Then $det(A) = det(B)det(C)$ where $B$ - upside-left matrix and $C$ downside-right matrix. $\endgroup$ – openspace Jan 25 '17 at 19:56
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Your matrix $A_n$ has the block diagonal structure

$$ A_n = \begin{pmatrix} B_n & 0 \\ 0 & B_n \end{pmatrix} $$

where $B_n \in M_n(\mathbb{F})$ is the matrix which has diagonal entries zero and all other entries $1$. Hence, $\det(A_n) = \det(B_n)^2$ so it is enough to calculate $\det(B_n)$. To do that, let $C_n$ be the matrix in which all the entries are $1$ (so $B_n = C_n - I_n$).

The matrix $C_n$ is a rank-one matrix so we can find it's eigenvalues easily. Let us assume for simplicity that $n \neq 0$ in $\mathbb{F}$. Then $C_n$ has an $n - 1$ dimensional kernel and $(1,\dots,1)^T$ is an eigenvalue of $C_n$ associated to the eigenvalue $n$. From here we see that the characteristic polynomial of $C_n$ must be $\det(\lambda I - C_n) = \lambda^{n-1}(\lambda - n)$ and hence $$\det(B_n) = \det(C_n - I_n) = (-1)^n \det(I_n - C_n) = (-1)^{n} 1^{n-1}(1 - n) = (-1)^n(1 - n) = (-1)^{n-1}(n-1).$$

In fact this formula works even if $n = 0$ in $\mathbb{F}$ because in this case, $A^2 = 0$ so $A$ is nilpotent and $\det(C_n - \lambda I) = \lambda^n$.

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As I commented : $detA = detB\cdot detc$, where $B = C$ with zero elements $a_{i,j}$ when $i = j$ and $1$ otherwise.

Now it's easy to see that $detB = n-1$, using definition of determinant. We have only one zero position in every row and only one $\sigma \in S_n$, which give us non-zero addition. So we get $det(B) = 1+\dots +1 = n - 1$.

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  • $\begingroup$ we have a lot of values of $\sigma$, every derrangement works. $\endgroup$ – Jorge Fernández Hidalgo Jan 25 '17 at 20:06
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The matrix is diagonal by blocks. so the determinant is equal to the product of the determinants of each block.

So we just have to prove that:

$\det\begin{pmatrix}0 & 1 & \dots & 1 \\ 1 & 0 & \dots & 1 \\ & &\vdots \\ 1 & 1 & \dots & 0\\ \end{pmatrix}=\pm(n-1)$

This is done in various ways here, in fact it is equal to $(-1)^n(n-1)$

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