# Rank of a $n! \times n$ matrix

This question is about showing that $n!$ points resulting from applying a function (defined below) to the permutations of $n$ numbers lie on a $n-1$ dimensional hyperplane.

Let $X=\langle x_1,\cdots,x_n \rangle$ and $Y=\langle y_1,\cdots,y_n \rangle$ be vectors of positive reals and let $P=\langle p_1,\cdots,p_n \rangle$ be a permutation of the numbers $\{1,\cdots,n\}$. Let,

$$y_i=\log\left(1+\frac{x_i}{1+\sum\limits_{p_j<\,p_i}x_j}\right)$$ be a mapping from $X,P$ to $Y$. So, for any given vector $X$, we have $n!$ (one for each permutation) output vectors $Y$. Let $Y_P$ denote a vector resulting from permutation $P$. Now, for a given $X$, create a matrix $M$ such that each row of it is $Y_P-Y_{P_1}$ for a fixed $P_1$ and for all $P$.

$$M=\begin{bmatrix} Y_{P_1}-Y_{P_1} \\ Y_{P_2}-Y_{P_1} \\ \vdots \\ Y_{P_{n!}}-Y_{P_1} \end{bmatrix}$$

So, $M$ is a $n! \times n$ matrix. I have two questions:

1. I guess the rank of $M$ is at most $n-1$, that is all of $Y$'s lie on a hyperplane of dimension $n-1$ (I verified it for $n=3$ and $n=4$). Is that true or not? Why?
2. For what sort of mappings from $X,P$ to $Y$, is the answer to the previous question positive?

Edit 3:

I realized that for all of the permutations, $\sum_i y_i = \log(1+\sum_i x_i)$ and as Chris Culter commented below, this might be the solution to the problem. The proof of inequality is long but an example clears its correctness: consider $n=3$ and $P=\langle 1, 2, 3 \rangle$,

\begin{align} y_1+y_2+y_3&= \log(1+\frac{x_1}{1+x_2+x_3})+ \log(1+\frac{x_2}{1+x_3})+ \log(1+\frac{x_3}{1})\\ &= \log(\frac{1+x_1+x_2+x_3}{1+x_2+x_3})+ \log(\frac{1+x_2+x_3}{1+x_3})+ \log(\frac{1+x_3}{1})\\ &= \log(1+x_1+x_2+x_3). \end{align}

Since any other permutation only renames $x_1$, $x_2$, and $x_3$, the output remains the same.

Edit 2:

The problem has a specific structure which may help solving it: The $n!$ points consist of $n$ ensembles of $(n-1)!$ points. Ensemble $i$ consists of all of the points $Y_P$ such that $i^{th}$ element of $P$ is 1. Therefore, all of the points in ensemble $i$ has the same $y_i$.

For example, for $n=3$, there are 3 ensembles of 2 points (6 points in total). All of the points lie on a 2D plane. The following figure shows an example in 3D:

Points in the same ensemble have a same colour. The labels beside the points denote the permutation that generates the point. So, points 123 and 132 are in ensemble 1, points 213 and 312 are in ensemble 2 (1 is the second element of their permutation), and points 231 and 321 are in ensemble 3 (1 is the third element of their permutation).

For $n=4$, there are 4 ensembles of 6 points (24 points in total). Since all of the points in 4D lie on a 3D plane, we can project the points into the 3D space. Here is an example:

Higher dimensions are recursively constructed as described above. So, for $n=5$, there are 5 ensembles of the form shown above (in 4D). This recursive nature can be useful for solving the problem.

Edit 3:

• I used Mathematica to verify my conjecture for $n=3$ to $8$ and it's correct. See my other question on Mathematica.SE. For $n>8$, because of the exponential running time of the algorithm, it's hard to verify the conjecture.

• The above conjecture also holds for the following mapping functions:

$$y_i={x_i}-{\sum\limits_{p_j<\,p_i}x_j}$$

$$y_i={x_i}^3-{\sum\limits_{p_j<\,p_i}{x_j}^2}$$

• But it does not hold for the following functions:

$$y_i=\log\left(\frac{x_i}{1+\sum\limits_{p_j<\,p_i}x_j}\right)$$

$$y_i={x_i}^3-\left(\sum\limits_{p_j<\,p_i}x_j\right)^2$$

• What is «it» in the second question? May 10, 2013 at 8:31
• The rank of a matrix can never be any higher than its smallest dimension-you might see this by considering the row reduction. May 10, 2013 at 8:43
• Note that $i$ is now a column index; rows are indexed by $p$. For $n=2$ I get $\begin{pmatrix} \log(1+x_1) & \log(1+\frac{x_2}{1+x_1}) \\ \log(1+x_1) & \log(1+\frac{x_2}{1+x_2}) \end{pmatrix}$ whose rank in general is $2$, not${}\leq1$. For instance for $x_1=1,x_2=2$ it gives $\begin{pmatrix} \log(2) & \log(2) \\ \log(2) & \log(5/3) \end{pmatrix}$ which is clearly full-rank. In other words your guess seems wrong. It would be good to show what you did actually check for $n=3$. May 10, 2013 at 9:39
• Hi there! I made the corrections and removed my comments. I also edited the question and added some new details and observations. I think the question is solvable now. Could you please have another look at it and give it another try. My feeling is that the answer is like this is obvious because ..., but I don't know the because part! May 19, 2013 at 4:32
• FWIW, the diagrams are reminiscent of permutohedra. Aug 22, 2013 at 0:54

Answer to the first question As Chris Culter guessed, the answer is YES and this is because $\sum_{i}y_i=\ln(1+\sum_{i} x_i)$. The proof of this last identity is not difficult, it is basically a change of indices and a careful use of notation (see below). What you have here is a telescoping sum (or product, depending on the way in which you look at it), but disposed according to a random order (depending on the permutation $p$).

$$\begin{array}{lcl} e^{\prod_{i=1}^n y_i} &=&\prod_{i=1}^n e^{y_i} \\ &=& \prod_{i=1}^n \frac{1+x_i+\sum_{p(j)<p(i)}x_j}{1+\sum_{p(j)<p(i)}x_j} \\ &=& \prod_{I=1}^n \frac{1+x_{p^{-1}(I)}+\sum_{J<I}x_{p^{-1}(J)}} {1+\sum_{J<I}x_{p^{-1}(J)}} \ \text{(where we put } I=p(i),J=p(j) \text{)} \\ &=& \prod_{I=1}^n \frac{1+z_I+\sum_{J=1}^{I-1}z_J} {1+\sum_{J=1}^{I-1}z_J} \ \text{(where we put } z_I=x_{p^{-1}(I)} \text{)} \\ &=& \prod_{I=1}^n \frac{t_I} {t_{I-1}} \ \text{(where we put } t_I=1+\sum_{J=1}^{I}z_J \text{)} \\ &=& \frac{t_N} {t_0}=1+x_1+x_2+\ldots +x_n. \end{array}$$

Partial answer to the second question

When $y_i$ is of the form $A(x_i)-\sum_{p(j)<p(i)}B(x_j)$ where $A$ and $B$ are functions, your conjecture still holds, because

$$\sum_{i}B(x_i)y_i=\sum_{i}A(x_i)B(x_i)-\frac{1}{2}\sum_{I\neq J}B(x_I)B(x_J)$$

When $A(x)=x$ and $B(x)=x$, you get $y_i=x_i-\sum_{p(j)<p(i)}x_j$.

When $A(x)=x^3$ and $B(x)=x^2$, you get $y_i=x_i^3-\sum_{p(j)<p(i)}x_j$.