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I have the matrix $A$ below which is the permutations of $\{f_0,f_1,f_2,f_3\}$

\begin{equation*} A=\begin{bmatrix} f_0 & f_1 & f_2 & f_3 \\ f_0 & f_1 & f_3 & f_2 \\ f_0 & f_2 & f_1 & f_3 \\ f_0 & f_2 & f_3 & f_1 \\ f_0 & f_3 & f_1 & f_2 \\ f_0 & f_3 & f_2 & f_1 \\ f_1 & f_0 & f_2 & f_3 \\ f_1 & f_0 & f_3 & f_2 \\ f_1 & f_2 & f_0 & f_3 \\ f_1 & f_2 & f_3 & f_0 \\ f_1 & f_3 & f_0 & f_2 \\ f_1 & f_3 & f_2 & f_0 \\ f_2 & f_0 & f_1 & f_3 \\ f_2 & f_0 & f_3 & f_1 \\ f_2 & f_1 & f_0 & f_3 \\ f_2 & f_1 & f_3 & f_0 \\ f_2 & f_3 & f_0 & f_1 \\ f_2 & f_3 & f_1 & f_0 \\ f_3 & f_0 & f_1 & f_2 \\ f_3 & f_0 & f_2 & f_1 \\ f_3 & f_1 & f_0 & f_2 \\ f_3 & f_1 & f_2 & f_0 \\ f_3 & f_2 & f_0 & f_1 \\ f_3 & f_2 & f_1 & f_0 \end{bmatrix}\end{equation*}

From visual inspection, the sixth row can be constructed from a linear combination of the preceding five rows (i.e.: $f_0$ in column 1 is in each of the five proceeding rows, $f_3$ in column 2 is in the fifth row, etc). Similarly, rows 8,11,12,14-18 and 20-24 can be constructed from linear combinations of preceding rows.

I am trying to figure out how to present the text in the paragraph immediately above in a concise mathematical way (i.e.: that these 14 rows are a linear combination of preceding rows, or conversely the other 10 rows that are completely independent.

HVW
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  • I'm very confused about what you're trying to say. The fact that each entry of the sixth row matches the corresponding entry of an earlier does not have any obvious connection to the fact that the sixth row is a linear combination of the previous rows. – Ben Grossmann Dec 27 '21 at 15:17
  • This matrix has 4 columns, so it's rank is at most 4. – mihaild Dec 27 '21 at 17:46

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For the sake of my sanity in trying to typeset this, I'll use $a,b,c,d$ instead of $f_0,f_1,f_2,f_3$. If you choose the four rows representing cyclic rotations $$R:=\begin{bmatrix}a&b&c&d\\d&a&b&c\\c&d&a&b\\b&c&d&a\end{bmatrix}$$ Then the determinant is $$|R| = (a^2 - c^2)^2 - (d^2-b^2)^2 - 4ac(d^2-b^2) - 4bd(a^2-c^2)$$

which is not $0$ in general (for instance, when $(a,b,c,d) = (0,1,2,3)$, the determinant is $-16$). Thus $R$ is invertible. You can find $R^{-1}$ using gaussian elimination, but I will not do so here because the result is messy. But because $R^{-1}$ exists, we have $$I = RR^{-1}$$ or $$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} = \begin{bmatrix}a&b&c&d\\d&a&b&c\\c&d&a&b\\b&c&d&a\end{bmatrix}R^{-1}$$ That is, there are linear combinations of these four rows in $R$ that give $$\begin{bmatrix}1&0&0&0\end{bmatrix},\begin{bmatrix}0&1&0&0\end{bmatrix},\begin{bmatrix}0&0&1&0\end{bmatrix},\begin{bmatrix}0&0&0&1\end{bmatrix}$$ And any four-dimensional row vector can be written as a linear combination of those four vectors, and thus in turn as a linear combination of the four rows in $R$.

Thus all rows of your matrix can be written as a linear combination of the four cyclic rotations.

Paul Sinclair
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