# Equality of determinants for a specific collection of square matrices of size $n=2^m$

My investigations have led me to a question that I am convinced is true. I need to show that, for a given $m$, a certain collection of square $n=2^m$ matrices have the same determinant. In dimension four, these matrices are $\left[ \begin{array}{cc} a_{11} & a_{12}&a_{13} &a_{14} \\ a_{21} & a_{22}&a_{23} &a_{24} \\ a_{31} & a_{32}&a_{33} &a_{34} \\ a_{41} & a_{42}&a_{43} &a_{44} \\ \end{array} \right]$, $\left[ \begin{array}{cc} a_{11} & -a_{12}&a_{13} &-a_{14} \\ -a_{21} & a_{22}&-a_{23} &a_{24} \\ a_{31} & -a_{32}&a_{33} &-a_{34} \\ -a_{41} & a_{42}&-a_{43} &a_{44} \\ \end{array} \right]$, $\left[ \begin{array}{cc} a_{11} & a_{12}&-a_{13} &-a_{14} \\ a_{21} & a_{22}&-a_{23} &-a_{24} \\ -a_{31} & -a_{32}&a_{33} &a_{34} \\ -a_{41} & -a_{42}&a_{43} &a_{44} \\ \end{array} \right]$, $\left[ \begin{array}{cc} a_{11} & -a_{12}&-a_{13} &a_{14} \\ -a_{21} & a_{22}&a_{23} &-a_{24} \\ -a_{31} & a_{32}&a_{33} &-a_{34} \\ a_{41} & -a_{42}&-a_{43} &a_{44} \\ \end{array} \right]$. In general the patterns can be generated by taking $\omega_q(p)$ to be $(-1)^{\lfloor \frac{p}{2^q} \rfloor}$, $p \in {0,1, \ldots, n-1 }$ and $q \in {0,1, \ldots, m }$. Then, in dimension $n=2^m$ and $A=\left( a_{rs} \right)$ we define $A_q$ to be $A_q=\left(\omega_q(r-1) \omega_q(s-1) a_{rs} \right)$. This generates $m+1$ of the $n$ matrices. The sign patterns for the rest of the matrices can be found by thinking of the ordered $n$-tuples $\left(\omega_q(0), \omega_q(1) , \ldots, \omega_q(n-1)\right)$ as generators of a multiplicative group where the multiplication takes place coordinate-wise. Thus $(1,-1,1,-1) \cdot (1,1,-1,-1) = (1,-1,-1,1)$ which explains the presence of the fourth matrix in the example above. If we arbitrarily assign these patterns to the $\omega_q(p)$, then we can get the remaining matrices, $A_q$, by defining them in the same way we did above. The question is just this: is it true that $\left| A_q \right| = \left| A_t \right|$ for all $q$ and $t$ between 0 and $n-1$? A reference to a proof would be ideal. Just a proof for the $4 \times 4$ case would be extremely helpful. Even if you know that (some if not all of) these matrices have a name, it would help. If it matters, the matrices I am considering are symmetric with entries that follow a particular pattern similar to the $\omega_q(p)$. I could describe these patterns if necessary, but I don’t think that they change the result of the question.

In other words, you multiply each row and column of the square matrix $A$ by the same $\omega_q$. Therefore the determinant changes by $\prod \omega_q^2 = 1$.
• @PaulSundheim The sign of column $i$ is not the same as that of row $i$? This is true for all of your examples and of your formula for $A_q$, unless I'm misunderstanding. Commented Aug 29, 2015 at 18:48
• @PaulSundheim But you are: you pick some subset of the rows, and multiple them all by $-1$. Then you pick the same subset of the columns, and multiply them by $-1$. Each multiplication flips the sign of the determinant. Commented Aug 29, 2015 at 19:08
• Put differently, $A_q = D_q A D_q$ where $D_q$ is a diagonal matrix of signs (the precise pattern of signs is unimportant). Commented Aug 29, 2015 at 19:12