How to deal with this system of equations?

Question

I am trying to investigate this system of equations:

$$ x^Tx-\mathbf{1}^T x=x^TP^tx=c $$

where $c\in\mathbb{Z}^+$ is a given positive integer, $\mathbf{1}=\{1,1,\ldots,1\}$ is a vector with as much ones as elements in $x$ and nothing else, $t\in\mathbb{Z}^+$ is any positive integer except if $P^t=I$ and

$$P=\begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots& \vdots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 &0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\ 1 & 0 & 0 & \cdots & 0 & 0 & 0 \end{pmatrix}$$

is a permutation matrix that shifts the elements of $x$ by one position at a time.

Using alternative notation we can write the same equations as $\sum x_ix_{i+t}=c$ where $i+t$ should be done $\mod n$ (number of components in $x$). and we have different equation for $t=0$: $\sum x_i^2 - \sum x_i=c$.

Question

I am familiar with basics of linear algebra, matrices etc. but I don't think I am familiar with systems like this one. I need a bit of guidance to get a grasp, anything of the following might be of a great help:

• How do we generally deal with equations $x^TAx=c$? And with $x^TA^tx=c$?
• Is there a special name for this system of equations?
• Is there any theory around such systems, what keywords should I search for?
• More exactly - how can I analyze this system, what can I learn about $x$ from these equations?

Some results

By summing all of the equations one can find that

$$ (\mathbf{1}^Tx)^2 - \mathbf{1}^Tx = n\lambda $$

It's useful to assign the root to $c'$: $$ c'= \mathbf{1}^Tx = (1\pm\sqrt{1+4nc})/2 $$

It appears (numerical evidence) that: $$\sum\limits_{\text{even }i} x_i = (c' \pm \sqrt{c'})/2$$ $$\sum\limits_{\text{odd }i} x_i = (c' \mp \sqrt{c'})/2$$

But I only managed to show that explicitly for $n=2$ and $n=4$. It is done by combining equations to eliminate anything but $x_1$ or $x_1+x_3$ respectively which than turns out to be described by a quadratic equation with the above roots.

Using similar approach I found that for $n=3$: $$ \frac{(c'- 2\sqrt{c'})}{3} \leq x_i \leq \frac{(c'+ 2\sqrt{c'})}{3} $$

Answer 1

Let confirm me if these equations are the correct representation of the constraints for the problem?. If you do we can move further into the closed solution.

In algebraic notation:

$$ \mathbf{P}_0: \sum_{i=1}^n x_i^2-x_i=c\\ \mathbf{P}_t,t:1...n-1: \sum_{i=1}^n x_ix_{i+j}=c\\ $$

Thus in matricial notation, we should have:

$$ \mathbf{P}_0: x^Tx-1^Tx=c\\ \mathbf{P}_t,t:1...n-1: x^TP_tx=c\\ $$

with the permutation matrix obtained from the identity or multiplying $t$ times:

$$ P_t=\left[I_{t...n}I_{1...n-1}\right]=P_1^t $$

and as general case:

$$ \mathbf{P}_t,t:1...n-1: x^TP_tx-\delta_{0t}1^Tx=c $$

The simplest $\mathbb{R}^2$ case, with $c=1$:

$$ 2xy=1\\ x^2+y^2-x-y=1 $$

leads to : $$ x=1/2 (2 \pm \sqrt 2)\\ y=1/2 (2 \mp \sqrt 2) $$

Here it is clear that $\mathbf{P}_0$ is unit nD-sphere 2D-surface centered in $x_i^0=1/2$ and with radius $r=\sqrt{c-n/4}$:

$$ \mathbf{P}_0: (x-x^0)^T(x-x^0)=c-n/4\\ $$

And $\mathbf{P}_t$ is nD-hyperboloid, 2D-surface, centered in the origin, and having $x_i=x_{i+t},i=1:n$ as the main axis.

Answer 2

A good area of mathematics to acquaint yourself with are those results dealing with the numerical range. The numerical range of the matrix $P^t$ is

$$\left\{\mathbf{y}^*P^t\mathbf{y} \mid \mathbf{y}\in\mathbb{C}^n,\ ||y||_2=1\right\},$$

which is a normalized version of the quadratic form that you're interested in. Note that $P^t$ in your case is a normal matrix, which implies that the numerical range of $P^t$ is the convex hull of the eigenvalues of $P^t$ (and therefore $P$). If a scalar $d\ne 0$ lies in the numerical range of $P^t$ for a particular unit vector $y,$ then you can construct a desired solution $x=\sqrt{\frac{c}{d}}y.$

Answer 3

I asked for references on the method used below at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr My preference is to write the steps as separate matrices and multiply them together; thus you find, below, r = r1 * r2 * r3 * r4 for example

There is a way to avoid eigenvalues and still get a diagonal quadratic form. It seems the eigenvalues are nice in even dimension but less nice in odd dimension. Double your thing reads $$ 2 x^T x - x^T(P + P^T)x - 2 1^T x = 2c. $$ The first point is that a quadratic form depends only on the symmetric part of the coefficient matrix.

$$ H = 2I - P^T - P. $$

Next, with a symmetric integer matrix, we can arrange a rational matrix of determinant $1,$ I called it $R,$ such that $$ R^T H R = D $$ is rational diagonal. Since $R$ is upper triangular of determinant $1,$ it is not such a big deal to find $R^{-1},$ which you will likely want.

If I get the patience I will typeset later, meanwhile there is a pattern, here is $n=5,$ below that $n=6.$

? h
%32 = 
[2 -1 0 0 -1]

[-1 2 -1 0 0]

[0 -1 2 -1 0]

[0 0 -1 2 -1]

[-1 0 0 -1 2]

? r = r1 * r2 * r3 * r4
%28 = 
[1 1/2 1/3 1/4 1]

[0 1 2/3 1/2 1]

[0 0 1 3/4 1]

[0 0 0 1 1]

[0 0 0 0 1]

? rt = mattranspose(r)
%29 = 
[1 0 0 0 0]

[1/2 1 0 0 0]

[1/3 2/3 1 0 0]

[1/4 1/2 3/4 1 0]

[1 1 1 1 1]

? d = rt * h * r
%31 = 
[2 0 0 0 0]

[0 3/2 0 0 0]

[0 0 4/3 0 0]

[0 0 0 5/4 0]

[0 0 0 0 0]

=======================================================

? r = r1 * r2 * r3 * r4 * r5
%25 = 
[1 1/2 1/3 1/4 1/5 1]

[0 1 2/3 1/2 2/5 1]

[0 0 1 3/4 3/5 1]

[0 0 0 1 4/5 1]

[0 0 0 0 1 1]

[0 0 0 0 0 1]

? rt = mattranspose(r)
%26 = 
[1 0 0 0 0 0]

[1/2 1 0 0 0 0]

[1/3 2/3 1 0 0 0]

[1/4 1/2 3/4 1 0 0]

[1/5 2/5 3/5 4/5 1 0]

[1 1 1 1 1 1]

? h
%27 = 
[2 -1 0 0 0 -1]

[-1 2 -1 0 0 0]

[0 -1 2 -1 0 0]

[0 0 -1 2 -1 0]

[0 0 0 -1 2 -1]

[-1 0 0 0 -1 2]

? d = rt * h * r
%28 = 
[2 0 0 0 0 0]

[0 3/2 0 0 0 0]

[0 0 4/3 0 0 0]

[0 0 0 5/4 0 0]

[0 0 0 0 6/5 0]

[0 0 0 0 0 0]

? 

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