invert one column before matrix multiplication and multiply result with unit vector = still same ranking, why?

Question

I am currently trying to find a mathematical proof of the following for a research paper. It's been quite long since i did higher mathematics and english is not my first language, so go easy on me if i got the terms wrong:

We have two matrices zxh and hxe that are multiplied to get the matrix zxe. This matrix in turn gets multiplied with a transposed unit vector 1xz to get the sum of each column in zxe. The result is the transposed vector 1xe. The values of this vector are ranked, which is the final result.

Values in zxh range from -1 to 1 and values in hxe range from 0 to 1.

Now i noticed the following:

If one column in zxh gets inverted (multiplied by -1) and at the same time if i switch the values in the corresponding row in hxe with their subtraction to 1, i get the same result in the ranking.

From what i observed the values in 1xe after the changes differ in a constant value from the ones before the changes. this constant is equal to the unit vector times the inverted column in zxh.

Example:

$$ zxh = \begin{pmatrix} 1 & -0,5 & 0 & -1 \\ 1 & 1 & 1 & 1 \\ -0,5 & 0 & 0,5 & 0 \end{pmatrix}; hxe = \begin{pmatrix} 1 & 0,5 & 0 \\ 0,5 & 1 & 0,5 \\ 1 & 0,5 & 0,5 \\ 1 & 1 & 1 \end{pmatrix} \\$$ The result of $1xz * zxh * hxe $ is $\begin{pmatrix} 3,25 & 2 & 1 \end{pmatrix}$

Now if we make the changes described above (let's take the first column and row):

$$ zxh = \begin{pmatrix} -1 & -0,5 & 0 & -1 \\ -1 & 1 & 1 & 1 \\ 0,5 & 0 & 0,5 & 0 \end{pmatrix}; hxe = \begin{pmatrix} 0 & 0,5 & 1 \\ 0,5 & 1 & 0,5 \\ 1 & 0,5 & 0,5 \\ 1 & 1 & 1 \end{pmatrix} $$

The result of $1xz * zxh * hxe $ is $\begin{pmatrix} 1,75 & 0,5 & -0,5 \end{pmatrix}$ which is exactly 1,5 less than the previous result.

I need to prove that the ranking before and after applying the changes, stays the same. The values in the vector 1xe are not relevant, as long as the ranking is equal.

Would someone please draw out the steps to prove this formally?

Answer 1

I found a more or less formal proof for the observed phenomenon through splitting up the individual elements in the matrix, similar to this: k is zxh; i is hxe; u is zxe; t is 1xe

Before: $$ u_{11}^v = k_{11}*i_{11} + k_{12}*i_{21} + k_{13}*i_{31} + ... $$ After: $$ u_{11}^n = -k_{11}*(1-i_{11}) + -k_{12}*(1-i_{21}) + -k_{13}*(1-i_{31}) + ... $$ Proof that the delta between those changes is constant: $$ \Delta_{vn} u_{11} = k_{11} + k_{12} + k_{13} ... = const. $$ and the delta for all elements in t is also constant $$ \Delta_{vn} t_{1} = g_1*\Delta_{vn}u_{11} + g_2*\Delta_{vn}u_{21} ... = const. $$

Therefore rank(t) is the same as before the changes

Answer 2

The phenomena you observe must have something to do with the way that the matrices in question were generated in the first place, because they do not hold in general. For instance, if $$ zhx = \pmatrix{1 & 0 \\ 0 & 1}\\ hxe = \pmatrix{1 & 1 \\ 0 & 1}\\ 1xz = \pmatrix{1 & 1}, $$ (I'm assuming that this is what you intended for $1xz$, because you say that multiplying by it gives column-sums), then we have \begin{align} 1xe &= 1xz ~zhx ~xhe \\ &= \pmatrix{1 & 1} \pmatrix{1 & 0 \\ 0 & 1}\pmatrix{1 & 1 \\ 0 & 1}\\ &= \pmatrix{1 & 2} \end{align} But if we multiply the first column of $zhx$ by $-1$, then the product becomes \begin{align} 1xe &= 1xz ~zhx ~xhe \\ &= \pmatrix{1 & 1} \pmatrix{-1 & 0 \\ 0 & 1}\pmatrix{1 & 1 \\ 0 & 1}\\ &= \pmatrix{-1 & 0} \end{align} i.e., your column sums do not remain invariant.

Similarly, if we replace the first row of $xhe$ by "one minus the first row", we get \begin{align} 1xe &= 1xz ~zhx ~xhe \\ &= \pmatrix{1 & 1} \pmatrix{1 & 0 \\ 0 & 1}\pmatrix{0 & 0 \\ 0 & 1}\\ &= \pmatrix{0 & 1} \end{align} which is again different. So the observations you make are false for most possible matrices --- there must be something special about the way your matrices are generated (or perhaps an error in your computations).

NB: I haven't shown that your properties don't hold for 'most possible matrices' -- I've merely given one example. But any small perturbation of that example will be another example, and in fact almost all small perturbations of an example where your properties hold will result in an example where both fail. (This requires proof, but I don't think it'll be very enlightening to include such a proof here.)