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

Solution 1:

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 $R$.