Determinant of a matrix of ones, whose anti diagonal elements are zero

Question

I'm trying to prove a formula I have constructed for the determinant of a general $n\times n $ real matrix $A$, given here in the case $n=5$: $$ A = \begin{bmatrix} 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \end{bmatrix}. $$ That is, the matrix containing all 1's apart from the anti-diagonal which consists of zeros.

Using a simple matlab code I've computed the determinants for the first few values of $n$, and have come to the formula $$ \det A = \begin{cases} -(n-1) \hspace{1em}\mbox{ if }\,\,\, n \equiv -1,0\mod 4, \\ n-1 \hspace{2.2em}\mbox{ otherwise,} \end{cases} $$ but I'm unsure where to start to prove this.

Answer 1

Let $J$ be the $n\times n$ matrix of all zeros except ones on the anti-diagonal and $e$ be the vector of all ones. Then $$ A=-J+ee^T. $$ Now apply the matrix determinant lemma. Note that $J^{-1}=J$.

Answer 2

Another way, which also give hints on some properties of the matrix, is as follows.

If $\mathbf U_n$ indicates the $n \times n$ matrix with all ones, and similarly $\mathbf J_n$ the antidiagonal matrix, i.e. the Exchange Matrix, then we can write $$ {\bf A}_n = {\bf U}_n - {\bf J}_n $$

Then take the matrix $\mathbf S_n$ which is a Lower Triangular matrix, with all ones in the main and lower diagonals, and with $0$ in the upper ones.
Its determinant is clearly one, and the inverse is also LT, and is the matrix with $1$ in the upper diagonal, $-1$ in the first lower, and zero elsewhere. $$ {\bf S}_n ^{\, - \,1} = {\bf I}_n - {\bf E}_n $$ which is easy to demonstrate, and where the meaning of ${\bf E}_n $ as Shift Matrix is clear.

Then apply the transformation $$ {\bf S}_n {\bf A}_n {\bf S}_n ^{\, - \,1} = {\bf S}_n \left( {{\bf U}_n - {\bf J}_n } \right){\bf S}_n ^{\, - \,1} = {\bf S}_n {\bf U}_n {\bf S}_n ^{\, - \,1} - {\bf S}_n {\bf J}_n {\bf S}_n ^{\, - \,1} $$ which means: sum progressively the rows then subtract the columns.

You get $$ {\bf S}_n {\bf U}_n {\bf S}_n ^{\, - \,1} = \left( {\matrix{ 0 & 0 & \cdots & 1 \cr 0 & 0 & \cdots & 2 \cr \vdots & \vdots & \ddots & \vdots \cr 0 & 0 & \cdots & n \cr } } \right)\quad \quad {\bf S}_n {\bf J}_n {\bf S}_n ^{\, - \,1} = \left( {\matrix{ 0 & \cdots & { - 1} & 1 \cr \vdots & {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} & \vdots & \vdots \cr { - 1} & \cdots & 0 & 1 \cr 0 & \cdots & 0 & 1 \cr } } \right) $$ subtract and you get a simple matrix.
To calculate the determinant, bring last column to first place (taking note of the sign) and then left multiply by ${\bf J}_n $ to put it upside-down.

Since $$ \left| {{\bf J}_n } \right| = \left( { - 1} \right)^{\,\left\lfloor {n/2} \right\rfloor } $$ you end with $$ \left| {{\bf A}_n } \right| = - \left( {n - 1} \right)\left( { - 1} \right)^{\,n + \left\lfloor {n/2} \right\rfloor } $$

Answer 3

This is easy to calculate by row reduction:

Add all rows to row 1.

$$\det(A) = \begin{vmatrix} 1 & 1 & ... & 1 & 0 \\ 1 & 1 & ... & 0 & 1 \\ ... & ... & ... & ... & ... \\ 1 & 0 & ... & 1 & 1 \\ 0 & 1 & ... & 1 & 1 \end{vmatrix}=\begin{vmatrix} n-1 & n-1 & ... & n-1 & n-1 \\ 1 & 1 & ... & 0 & 1 \\ ... & ... & ... & ... & ... \\ 1 & 0 & ... & 1 & 1 \\ 0 & 1 & ... & 1 & 1 \end{vmatrix}\\=(n-1)\begin{vmatrix} 1 & 1 & ... & 1 & 1 \\ 1 & 1 & ... & 0 & 1 \\ ... & ... & ... & ... & ... \\ 1 & 0 & ... & 1 & 1 \\ 0 & 1 & ... & 1 & 1 \end{vmatrix}=(n-1)\begin{vmatrix} 1 & 1 & ... & 1 & 1 \\ 0 & 0 & ... & -1 & 0 \\ ... & ... & ... & ... & ... \\ 0 & -1 & ... & 0 & 0 \\ -1 & 0 & ... & 0 & 0 \end{vmatrix}$$ where in the last row we subtracted row 1 from all rows. Now add again all rows to row 1: $$\det(A)=(n-1)\begin{vmatrix} 0 & 0 & ... & 0 & 1 \\ 0 & 0 & ... & -1 & 0 \\ ... & ... & ... & ... & ... \\ 0 & -1 & ... & 0 & 0 \\ -1 & 0 & ... & 0 & 0 \end{vmatrix}$$

The last determinant is easy now to calculate.