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Consider the set of coordinates $X_{i, j}^{(\ell)}$ and $Y_{i, j}^{(\ell)}$, where $i \in (1, 2, 3), j \in (1, 2, 3)$ and $i \neq j$ and $\ell = \pm 1$. The superscipt $(\ell)$ is an index. Consider the change of variables from $\mathbf{X}$ to $\mathbf{Y}$ defined by \begin{equation} Y_{i, j}^{(\ell)} = c_{i, j}^{(\ell)} + c_{i, k}^{(\ell)} X_{k, j}^{(\ell)} + c_{i, j}^{(-\ell)} X_{j, i}^{(-\ell)} X_{i, j}^{(\ell)} + c_{i, k}^{(-\ell)} X_{k, i}^{(-\ell)} X_{i, j}^{(\ell)} \tag{1} \end{equation} where $(i, j, k)$ is any permutation of $(1, 2, 3)$ and the $c_{i, j}^{(\ell)}$ are any non-zero numbers constrained, for each $i$, by $c_{i, j}^{(1)} + c_{i, k}^{(1)} + c_{i, j}^{(-1)} + c_{i, k}^{(-1)} = K$ for some constant $K > 0$. So, for example, if $i = 1$ then $c_{1, 2}^{(1)} + c_{1, 3}^{(1)} + c_{1, 2}^{(-1)} + c_{1, 3}^{(-1)} = K$.

Let $\mathbf{J}(\mathbf{X})= \partial \mathbf{Y}/ \partial \mathbf{X}$ be the Jacobian matrix of the transformation in Eq. $(1)$. Let $\mathbf{1}$ be a vector of length $12$ consisting of all $1$'s.

Fact 1: The constant $K$ is an eigenvalue of $\mathbf{J}(\mathbf{1})$.

Fact 2: If there is a vector $\mathbf{v}$ satisfying $\mathbf{J}(\mathbf{1}) \mathbf{v} = K \mathbf{v}$, then it has the property that $v_{i, j}^{(k)} + v_{j, i}^{(k)} = 0$, where $v_{a, b}^{(c)}$ is the entry of $\mathbf{v}$ corresponding to coordinate $X_{a, b}^{(c)}$.

I know Fact 1 and Fact 2 are facts, because the eigensystem of a $12 \times 12$ matrix is still (barely) within the realm of tractability for computers. My question is therefore:

Is there any way to see that either Fact 1 or Fact 2 is true without having to actually find the eigenvector?

A relevant, similar question I have asked is here. In that case, it was "easy" to guess the eigenvector but I can't see how to do that here. It has to be something about the structure of Eq. $(1)$. There's this idea that if the sum of the coefficients of a transformation is constrained to be a constant in the way I've described, then this constraint shows up as an eigenvalue of the Jacobian and I wonder how general this property is.

Mathematica Code

Using the convention $c_{i, j}^{(\ell)} = c[i, j, \ell]$, the Jacobian is

jac = {{c[1, 2, 1] + c[1, 3, 1], 0, 0, 0, 0, c[1, 2, 1], 0, 0, 0, c[1, 3, 1], c[1, 3, -1], 0},
{0, c[1, 2, -1] + c[1, 3, -1], 0, 0, c[1, 2, -1], 0, 0, 0, c[1, 3, -1], 0, 0, c[1, 3, 1]},
{0, 0, c[1, 2, 1] + c[1, 3, 1], 0, 0, c[1, 2, 1], c[1, 2, -1], 0, 0, c[1, 3, 1], 0, 0},
{0, 0, 0, c[1, 2, -1] + c[1, 3, -1], c[1, 2, -1], 0, 0, c[1, 2, 1], c[1, 3, -1], 0, 0, 0},
{0, c[2, 1, 1], 0, 0, c[2, 1, 1] + c[2, 3, 1], 0, 0, 0, c[2, 3, -1], 0, 0, c[2, 3, 1]},
{c[2, 1, -1], 0, 0, 0, 0, c[2, 1, -1] + c[2, 3, -1], 0, 0, 0, c[2, 3, 1], c[2, 3, -1], 0},
{0, c[2, 1, 1], c[2, 1, -1], 0, 0, 0, c[2, 1, 1] + c[2, 3, 1], 0, 0, 0, 0, c[2, 3, 1]},
{c[2, 1, -1], 0, 0, c[2, 1, 1], 0, 0, 0, c[2, 1, -1] + c[2, 3, -1], 0, 0, c[2, 3, -1], 0},
{0, 0, 0, c[3, 1, 1], c[3, 2, -1], 0, 0, c[3, 2, 1], c[3, 1, 1] + c[3, 2, 1], 0, 0, 0},
{0, 0, c[3, 1, -1], 0, 0, c[3, 2, 1], c[3, 2, -1], 0, 0, c[3, 1, -1] + c[3, 2, -1], 0, 0},
{c[3, 1, -1], 0, 0, c[3, 1, 1], 0, 0, 0, c[3, 2, 1], 0, 0, c[3, 1, 1] + c[3, 2, 1], 0},
{0, c[3, 1, 1], c[3, 1, -1], 0, 0, 0, c[3, 2, -1], 0, 0, 0, 0, c[3, 1, -1] + c[3, 2, -1]}}

Where the rows are in the order

rows = {c[1, 2, -1], c[1, 2, 1], c[1, 3, -1], c[1, 3, 1], c[2, 1, -1], c[2, 1, 1],
      c[2, 3, -1], c[2, 3, 1], c[3, 1, -1], c[3, 1, 1], c[3, 2, -1], c[3, 2, 1]}

Here's a sample set of $c_{i,j}^{(\ell)}$ where $K = 1$:

vec = {0.213509, -0.0587312, 0.321703, 0.523519, 0.255562, -0.290697,
     0.211956, 0.823179, 0.0321628, 1.087800, -0.373663, 0.253700}

Running the following command:

Eigenvalues[jac /. Thread[rows -> vec]]

Shows that $1$ is indeed an eigenvalue.

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Let $M=\mathbf{J(1)}\color{red}{-KI_{12}}$ and \begin{aligned} v&=(v_1,v_2,\ldots,v_{12})^T\\ &=\left(v_{1,2}^{(-1)},\ v_{1,2}^{(1)},\ v_{1,3}^{(-1)},\ v_{1,3}^{(1)},\ v_{2,1}^{(-1)},\ v_{2,1}^{(1)},\ v_{2,3}^{(-1)},\ v_{2,3}^{(1)},\ v_{3,1}^{(-1)},\ v_{3,1}^{(1)},\ v_{3,2}^{(-1)},\ v_{3,2}^{(1)},\ \right)^T. \end{aligned} Then $\mathbf{J(1)}v=Kv$ if and only if $Mv=0$.

Numerical experiments suggest that the nullspace of $M$ contains vectors $v$ with $$ v_1=-v_5,\quad v_2=-v_6,\quad v_3=-v_9,\quad v_4=-v_{10},\quad v_7=-v_{11},\quad v_8=-v_{12}. $$ Therefore, to prove both fact 1 and fact 2, it suffices to show that the following $12\times6$ matrix is not of full column rank: $$ P=\pmatrix{m_{\ast1}-m_{\ast5}&m_{\ast2}-m_{\ast6}&m_{\ast3}-m_{\ast9}&m_{\ast4}-m_{\ast10}&m_{\ast7}-m_{\ast11}&m_{\ast8}-m_{\ast12}}.\tag{1} $$ Using the value of $\mathbf{J(1)}$ given by your Mathematica output, we obtain $$ P=\pmatrix{ -x_3-x_4 &-x_1 &0 &-x_2 &-x_4 &0\\ -x_3 &-x_1-x_2 &-x_4 &0 &0 &-x_2\\ 0 &-x_1 &-x_3-x_4 &-x_2 &x_3 &0\\ -x_3 &0 &-x_4 &-x_1-x_2 &0 &x_1\\ x_7+x_8 &x_6 &-x_7 &0 &0 &-x_5\\ x_8 &x_5+x_6 &0 &-x_5 &-x_7 &0\\ 0 &x_6 &x_8 &0 &-x_7-x_8 &-x_5\\ x_8 &0 &0 &x_6 &-x_7 &-x_5-x_6\\ -x_{12} &0 &x_{11}+x_{12} &x_9 &0 &x_{10}\\ 0 &-x_{10} &x_{11} &x_9+x_{10}&x_{12} &0\\ x_{11} &0 &0 &x_9 &x_{11}+x_{12} &x_{10}\\ 0 &x_9 &x_{11} &0 &x_{12} &x_9+x_{10}} $$ where the $x_n$s are shorthands for the $X_{i,j}^{(\ell)}$s: $$ \pmatrix{ x_1&x_2&x_3&x_4\\ x_5&x_6&x_7&x_8\\ x_9&x_{10}&x_{11}&x_{12}} =\pmatrix{ X_{1,2}^{(-1)} &X_{1,2}^{(1)} &X_{1,3}^{(-1)} &X_{1,3}^{(1)}\\ X_{2,1}^{(-1)} &X_{2,1}^{(1)} &X_{2,3}^{(-1)} &X_{2,3}^{(1)}\\ X_{3,1}^{(-1)} &X_{3,1}^{(1)} &X_{3,2}^{(-1)} &X_{3,2}^{(1)}}. $$ Now perform the following column operations on $P$: $$ C1\leftarrow C1-C5,\quad C2\leftarrow C2-C6,\quad C3\leftarrow C3+C5,\quad C4\leftarrow C4+C6.\tag{2} $$ The submatrix containing the first four columns of matrix $P$ then becomes: $$ Q=\pmatrix{ -x_3 &-x_1 &-x_4 &-x_2 \\ -x_3 &-x_1 &-x_4 &-x_2 \\ -x_3 &-x_1 &-x_4 &-x_2 \\ -x_3 &-x_1 &-x_4 &-x_2 \\ x_7+x_8 &x_5+x_6 &-x_7 &-x_5 \\ x_7+x_8 &x_5+x_6 &-x_7 &-x_5 \\ x_7+x_8 &x_5+x_6 &-x_7 &-x_5 \\ x_7+x_8 &x_5+x_6 &-x_7 &-x_5 \\ -x_{12} &-x_{10} &x_{11}+x_{12} &x_9+x_{10}\\ -x_{12} &-x_{10} &x_{11}+x_{12} &x_9+x_{10}\\ -x_{12} &-x_{10} &x_{11}+x_{12} &x_9+x_{10}\\ -x_{12} &-x_{10} &x_{11}+x_{12} &x_9+x_{10}}. $$ Since the row rank of $Q$ is $\le3$, $Q$ has not full column rank. Thus $Q(x,y,z,w)^T=0$ for some $(x,y,z,w)\ne0$. By $(2)$, this means $Pu=0$ when $$ u=x(e_1'-e_5')+y(e_2'-e_6')+z(e_3'+e_5')+w(e_4'+e_6') =(x,\ y,\ z,\ w,\ z-x,\ w-y)^T $$ where $\{e_1',e_2',\ldots,e_6'\}$ denotes the standard basis of $\mathbb R^6$. In turn, by $(1)$, we have $Mv=0$ (or equivalently, $\mathbf{J(1)}v=Kv$) when \begin{aligned} v&=\left(v_{1,2}^{(-1)},\ v_{1,2}^{(1)},\ v_{1,3}^{(-1)},\ v_{1,3}^{(1)},\ v_{2,1}^{(-1)},\ v_{2,1}^{(1)},\ v_{2,3}^{(-1)},\ v_{2,3}^{(1)},\ v_{3,1}^{(-1)},\ v_{3,1}^{(1)},\ v_{3,2}^{(-1)},\ v_{3,2}^{(1)}\right)^T\\ &=u_1(e_1-e_5)+u_2(e_2-e_6)+u_3(e_3-e_9)+u_4(e_4-e_{10})+u_5(e_7-e_{11})+u_6(e_8-e_{12})\\ &=(x,\ y,\ z,\ w,\ -x,\ -y,\ z-x,\ w-y,\ -z,\ -w,\ x-z,\ y-w)^T. \end{aligned}

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  • $\begingroup$ Is there any underlying reason for the column operations? I noticed recently that the eigenvector also has the property that $$ \begin{align} v_{1, 3}^{(1)} &= v_{1, 2}^{(1)} + v_{2, 3}^{(1)} \\ v_{1, 3}^{(-1)} &= v_{1, 2}^{(-1)} + v_{2, 3}^{(-1)} \\ v_{3, 1}^{(1)} &= v_{3, 2}^{(1)} + v_{2, 1}^{(1)} \\ v_{3, 1}^{(-1)} &= v_{3, 2}^{(-1)} + v_{2, 1}^{(-1)} \end{align} $$ Is it related? $\endgroup$
    – Mr. G
    Jul 11 '19 at 23:20
  • $\begingroup$ @Mr.G You can get this by back-substituting the solution $r$ of $Qr=0$ into the solution $u$ of $Pu=0$ and then by back-substituting $u$ into the solution $v$ of $Mv=0$. See my new edit. $\endgroup$
    – user1551
    Jul 12 '19 at 3:28

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