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Background

Let $A = \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]$ be a matrix in $\mathbb{R}^{2 \times 2}$. While matrices are often used to represent a variety of linear transformations, including rotations, here I am transforming the matrix itself.

A classic way to introduce group transformations is via the following 8 transformations of a square with distinct corners:

  • Rotation of 0 degrees: $R_0(A)=\left[\begin{matrix}a & b\\c & d\end{matrix}\right]$
  • Rotation of 90 degrees: $R_{90}(A)=\left[\begin{matrix}b & d\\a & c\end{matrix}\right]$
  • Rotation of 180 degrees: $R_{180}(A)=\left[\begin{matrix}d & c\\b & a\end{matrix}\right]$
  • Rotation of 270 degrees: $R_{270}(A)=\left[\begin{matrix}c & a\\d & b\end{matrix}\right]$
  • Flipping about a horizontal axis: $H(A)=\left[\begin{matrix}c & d\\a & b\end{matrix}\right]$
  • Flipping about a vertical axis: $V(A)=\left[\begin{matrix}b & a\\d & c\end{matrix}\right]$
  • Flipping about the main diagonal: $D(A)=\left[\begin{matrix}a & c\\b & d\end{matrix}\right]$
  • Flipping about the other diagonal: $D^\prime(A)= \left[\begin{matrix}d & b\\c & a\end{matrix}\right]$

Considering the compositions of all pairs of operations yields an 8 by 8 Cayley table. Thinking of my matrix $A$ as the square to be rotated, I created a similar table except the entries are the determinants of the resulting matrices after applying a composition of two of the above operations which I have represented with matrix $B$ below. The order of the rows and columns follow the order of the transformations as they are listed above.

$$B = \left[\begin{matrix}a d - b c & - a d + b c & a d - b c & - a d + b c & - a d + b c & - a d + b c & a d - b c & a d - b c\\- a d + b c & a d - b c & - a d + b c & a d - b c & a d - b c & a d - b c & - a d + b c & a d - b c\\a d - b c & - a d + b c & a d - b c & - a d + b c & - a d + b c & - a d + b c & a d - b c & a d - b c\\- a d + b c & a d - b c & - a d + b c & a d - b c & a d - b c & a d - b c & - a d + b c & a d - b c\\- a d + b c & a d - b c & - a d + b c & a d - b c & a d - b c & a d - b c & - a d + b c & a d - b c\\- a d + b c & a d - b c & - a d + b c & a d - b c & a d - b c & a d - b c & - a d + b c & a d - b c\\a d - b c & - a d + b c & a d - b c & - a d + b c & - a d + b c & - a d + b c & a d - b c & a d - b c\\a d - b c & - a d + b c & a d - b c & - a d + b c & - a d + b c & - a d + b c & a d - b c & a d - b c\end{matrix}\right]$$

Finally I computed the determinant of $B$ to be $\det (B) = 0$.

Question

Just as I started with 2 by 2 matrices and got a final "determinant of the matrix of determinants of the transformations of the group applied to the original matrix", will I also such a determinant of zero for any n by n matrix?

Edits

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Starting with $A = \left[\begin{matrix}x_{0} & x_{1} & x_{2}\\x_{3} & x_{4} & x_{5}\\x_{6} & x_{7} & x_{8}\end{matrix}\right]$, I found that the matrix of determinants was

$$B = \left[\begin{matrix}x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6}\\- x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6}\\x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6}\\- x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6}\\- x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6}\\- x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6}\\x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6}\\x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & - x_{0} x_{4} x_{8} + x_{0} x_{5} x_{7} + x_{1} x_{3} x_{8} - x_{1} x_{5} x_{6} - x_{2} x_{3} x_{7} + x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6} & x_{0} x_{4} x_{8} - x_{0} x_{5} x_{7} - x_{1} x_{3} x_{8} + x_{1} x_{5} x_{6} + x_{2} x_{3} x_{7} - x_{2} x_{4} x_{6}\end{matrix}\right]$$

And I similarly found $\det(B) = 0$ as was found in the 2 by 2 case.

2

I found the resulting determinant be zero in the 4 by 4 and 5 by 5 cases. Note that the transformations above are rotating/reflecting the matrix entries themselves, rather than vectors in a vector space.

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  • $\begingroup$ Getting any rotations etc. is just the same as multiplying matrix with a proper unitary matrix. Since determinant of any unitary matrix is 1 or -1 and since $det(A*B) = det(A)*det(B)$ the resulting matrix has as its entries determinant of the original matrix multiplied by 1 or -1. So, the whole problem reduces to determining determinant (pfffu) of this 1, and -1 matrix. It is unclear to me how you list rows/columns of your matrix however ... $\endgroup$
    – Salcio
    Jan 18 at 2:51

2 Answers 2

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$\DeclareMathOperator\id{id}$ Let $A$ be an arbitrary $n \times n$ matrix. For each element $\sigma \in D_{8},$ we let $\sigma \cdot A$ denote the action of $\sigma$ on $A$ (so, if $\sigma$ is the rotation of the square by $90$ degrees counterclockwise, $\sigma \cdot A$ is the matrix obtained by rotating $A$ $90$ degrees counterclockwise).

Now, let $B$ be the matrix whose columns are labelled left to right with the elements of $D_{8}$ in some order (lets say $\sigma_{1}, \sigma_{2}, \ldots, \sigma_{8}$), and whose rows are labelled top to bottom with the elements of $D_{8}$ in the same order. If $\sigma, \sigma'$ are two elements of $D_{8},$ the entry in the $\sigma$ row and $\sigma'$ column of $B$ is $$\det(\sigma' \cdot (\sigma \cdot A)).$$

Note that rotation by $0$ degrees (the identity) and reflection across the main diagonal are elements of $D_{8}$; we let $\id$ and $\varphi$ denote these two, respectively. Furthermore, note that $\id \cdot A = A,$ while $\varphi \cdot A = A^{T},$ the transpose of $A$. We see that $\det(\id \cdot A) = \det(A)$, and that $\det(\varphi \cdot A) = \det(A^{T}) = \det(A).$

What are the entries of the $\id$ column of $B$? Well, for each $\sigma_{i} \in D_{8},$ we have $$\det(\id \cdot (\sigma \cdot A)) = \det(\sigma_{i} \cdot A),$$ so the entries of the $\id$ column of $B$ are just the determinants $\det(\sigma_{i} \cdot A),$ for $i \in [0, 8].$

Similarly, for each $\sigma_{i} \in D_{8},$ we have $$\det(\varphi \cdot (\sigma \cdot A)) = \det((\sigma_{i} \cdot A)^{T}) = \det(\sigma_{i} \cdot A),$$ so the entries of the $\varphi$ column of $B$ are also just the determinants $\det(\sigma_{i} \cdot A),$ for $i \in [0, 8].$

So, we see that $B$ has two identical columns, hence $\det(B) = 0.$ Note that this is true regardless of the size of $A$.

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  • $\begingroup$ Have you seen my answer ? $\endgroup$
    – Jean Marie
    Jan 18 at 11:22
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In fact, the operations described by @ckefa can be given an all-matricial form, providing a direct access to their determinant. I am going to present them in $n=2$ dimensions for the benefit of simplicity, but they are valid in any dimension $n$:

Let

$$J=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \ \ A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

Please note that anti-unit matrix $J$ has determinant $-1$ (valid in any dimension $n$).

The different operations you have can be given an all-matricial form ; using your notations, we have in particular:

  • direct (= anti-clockwise) rotation by 90° is $A \to R_{90}(M):=\color{red}{JA^T}=\begin{pmatrix}b&d\\a&c\end{pmatrix}.$

  • clockwise rotation by 90° is $A \to R_{90}^{-1}(A):=\color{red}{A^TJ}.$

  • symmetry wrt to a vertical axis (= placing the columns in reverse order) is $A \to V(A):=\color{red}{AJ}.$

  • symmetry wrt to a horizontal axis (= placing the lines in reverse order) is $A \to H(A):=\color{red}{JA}.$

Now, your matrix of determinants is more easily computable...

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  • $\begingroup$ For the "mattress group" itself, see this nice article bit-player.org/wp-content/extras/bph-publications/… $\endgroup$
    – Jean Marie
    Jan 18 at 20:54
  • $\begingroup$ Besides, for the Cayley table of the group, do you know that the factorisation of its determinant has led to an important theory called the theory of group characters ? See here $\endgroup$
    – Jean Marie
    Jan 18 at 21:05

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