Mathematical representation of row independence for a matrix that is not full rank 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.
 A: 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.
