Eigenvalues of the Jacobian matrix of a quadratic change of variables

Question

Consider the set of coordinates $x_i$ and $y_i$ for $i = \pm 1, \pm 2 \dots \pm N$ and $N \geq 2$. Consider the change of coordinates from $\mathbf{x}$ to $\mathbf{y}$ defined by $$ y_i(\mathbf{x}) = c_i + x_i \sum_{j \neq i} c_j x_{-j} \tag{1} $$ where the $c_i$ are any strictly positive real numbers constrained by $\sum_{i} c_i = K$ for a constant $K$. Let $\mathbf{J}(\mathbf{x}) = \partial \mathbf{y}/\partial \mathbf{x}$ be the Jacobian matrix of the transformation. Let $\mathbf{1}$ be a vector of length $2N$ consisting of all $1$'s.

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

Conjecture 2: Assuming Conjecture $1$ is true, any vector $\mathbf{v}$ satisfying $\mathbf{J}(\mathbf{1}) \mathbf{v} = K \mathbf{v}$ has the property that $v_{j} + v_{-j} = 0$, where $v_{\ell}$ is the entry of $\mathbf{v}$ corresponding to coordinate $x_\ell$.

I've verified both conjectures by explicit (computer-assisted) calculation for $N = 2, 3, 4$. One observation is that Eq. $(1)$ can be written as

\begin{align} y_i &= c_i + x_i \sum_{j} c_j x_{-j} - c_i x_{i} x_{-i} \\ &= c_i + \alpha x_i - c_i x_{i} x_{-i} \end{align} where $\alpha$ is a constant that doesn't depend on $i$. In particular, swapping $i \to -i$ shows that $y_i$ and $y_{-i}$ both (sort of) only depend on $x_i$ and $x_{-i}$ which is suggestive of Conjecture $2$. It really seems like there should be a simple reason for the conjectures but I can't see it.

Answer 1

I made a little computational mistake earlier. The underlying argument remains the same. Here is now the complete proof of both conjectures. The conjecture is equivalent to saying that for $n\ge2$ and $c_i\ne0,\,\forall i\in\{\pm1,\pm 2,\cdots,\pm n\}$ the column matrix $[v_i]_{{i=-n}\atop{i\ne0}}^n \ni \big(0=\sum_{j=1}^n (c_{-j}-c_j)v_j \bigwedge v_{-j}=-v_j,\,\forall j\ge1\big)$ is necessary and sufficient for $\mathbf J(\mathbf1)v=Kv$. The solution space of the first equation is $n-1$ dimensional. Note that $c_i$'s do not have to be positive but so long as they are nonzero.

Proof:

Using the Kronecker delta $\delta_{k,i}$, write $$y_i = c_i+x_i\sum_k(1-\delta_{k,i})c_kx_{-k}.$$ Compute the $(i,j)$'th entry of the Jacobian $\mathbf J(x)$ \begin{align} \mathbf J(x)_{i,j} &=\frac{\partial y_i}{\partial x_j} \\ &= \delta_{i,j}\sum_k(1-\delta_{k,i})c_kx_{-k}+x_i\sum_k(1-\delta_{k,i})c_k\delta_{-k,j} \\ &=\delta_{i,j}\sum_kc_kx_{-k}-c_i(\delta_{i,j}x_{-i}+\delta_{i,-j}x_i)+c_{-j}x_i. \end{align} Obviously $$\mathbf J(\mathbf 1)_{i,j} = K\delta_{i,j}-c_i(\delta_{i,j}+\delta_{i,-j})+c_{-j} \tag1$$ In the explicit matrix form $$\mathbf J(\mathbf 1)=KI-\text{diag}(c)-\text{adiag}(c)+\mathbf 1\tilde c^T,$$ where $c=[c_{-n},\cdots,c_{-1},c_1,\cdots,c_n]^T$ while $\tilde c=[c_n,\cdots,c_1,c_{-1},\cdots,c_{-n}]^T$, diag$(c)$ is the diagonal matrix with $c$ laying on its main diagonal, and adiag$(c)$ is the diagonal matrix with $c$ laying on its main anti-diagonal.

It is however more convenient to use expression $(1)$.

For $v$ to be an eigenvector of $\mathbf J(\mathbf 1)$ with eigenvalue $K$ or $$Kv_i=\sum_j\mathbf J(\mathbf 1)_{i,j}v_j=Kv_i-c_i(v_i+v_{-i})+\sum_jc_{-j}v_j$$ which is obviously equivalent to $$c_i(v_i+v_{-i}) = \sum_{j=1}^n (c_{-j}v_j+c_jv_{-j}), \,\forall i. \tag2$$ This means the left hand side of the above equation is independent of $i$.

1. Sufficiency or Conjecture 1).

It suffices to find a nonzero column matrix $[v_i]_{i=1}^n \ni 0=\sum_{j=1}^n (c_{-j}-c_j)v_j$ then let $v_{-j}=-v_j,\,\forall j\ge1$ and the solution space is $n-1$ or $n$ (if $c_{-j}-c_j,\,\forall i$) dimensional.

2. Necessity or Conjecture 2).

From Equation (2), $$c_i(v_i+v_{-i}) = c_{-i}(v_{-i}+v_i) \, \Longleftrightarrow\, (c_i-c_{-i})(v_i+v_{-i})=0,\,\forall i.$$ Then by Equation (2) again, $$\big(\exists k\ni c_k\ne c_{-k} \implies v_k+v_{-k}=0 \big)\implies \big(v_i+v_{-i}=0,\forall i \bigwedge \sum_{j=1}^n (c_{-j}-c_j)v_j=0\big).$$ Clearly $$c_i=c_{-i}\ne0\,\forall i \implies \sum_{j=1}^n (c_{-j}v_j+c_jv_{-j})=\sum_{j=1}^n c_j(v_j+v_{-j}).$$ But by Equation $(2)$, $c_j(v_j+v_{-j})$ are all the same for all $j$'s, Equation $(2)$ becomes $$c_i(v_i+v_{-i}) = nc_i(v_i+v_{-i}),\,\forall i\Longleftrightarrow (nc_i-1)(v_i+v_{-i})=0,\,\forall i.$$ Further more, \begin{align} \exists k\ni nc_k\ne1 &\implies v_k+v_{-k}=0 \\ &\implies c_i\ne0\wedge v_i+v_{-i}=0,\,\forall i; \end{align} and \begin{align} nc_i=1\,\forall i&\implies v_i+v_{-i}=n(v_i+v_{-i}) \\ &\implies n>1\wedge v_i+v_{-i}=0,\,\forall i. \end{align}

In conclusion $v_i+v_{-i}=0,\,\forall i$ so long as $n>1$ and $c_i\ne0$ if $c_i=c_{-i}$, $\forall i$. By Equation $(2)$, $$\sum_{j=1}^n (c_{-j}-c_j)v_j=0.$$

For $n=1$ and $c_1=c_{-1}=1$, $\mathbf J(\mathbf 1)=K\mathbf I$. The only eigenvalue is $K$ and any vector $2$ dimensional vector is an eigenvector.

Answer 2

Part 1: Conjecture 1 is true.

To simplify a little, first, I replace $y_i$ with \begin{align} y_i(\mathbf{x}) &= x_i \sum_{j \neq i} c_j x_{-j} \tag{1}\\ &= x_i \sum_{j } c_j x_{-j} - c_i x_i x_{-i} \tag{2} \end{align} (i.e., ignore the translation) because the translation has no effect on the Jacobian. I'll use formula 2 henceforth.

Computing the individual entries of the jacobian is straightforward. I'm going to use the differential geometer's "comma" notation, so that $y_{a,b}$ denotes

$$ \partial y_a/\partial x_b $$

So let's look at $y_{i,k}$; there are three cases: $k = i, k = -i,$ and $k \ne \pm i$. \begin{align} y_{i,i} &= \partial y_i/\partial x_i \\ &= \sum_{j } c_j x_{-j} + x_i c_{-i} - c_i x_{-i}\\ y_{i,-i} &= 0 \\ y_{i,k} &= x_i c_{-k} & \text{$k \ne i, -i$} \end{align}

Now evaluating this at $x = \mathbf{1}$, we get \begin{align} y_{i,i} &= \partial y_i/\partial x_i \\ &= K + c_{-i} - c_i \\ y_{i,-i} &= 0 \\ y_{i,k} &= c_{-k} & \text{$k \ne i, -i$} \end{align}

OK. Now let's look at the product of this Jacobian with a vector $v$; indeed, let's look at $Jv - Kv$, and see what it means to set this to zero. Let's find the $i$th entry of $Jv$: \begin{align} (Jv)_i &= \sum_k y_{i,k}v_k \\ &=\left( \sum_{k\ne \pm i} y_{i,k}v_k \right) + y_{i,i}v_i + y_{i,-i}v_{-i} \\ &=\left( \sum_{k\ne \pm i} c_{-k}v_k \right) + (K + c_{-i} - c_i)v_i + 0 \cdot v_{-i} \\ &=\left[ \left( \sum_{k} c_{-k}v_k \right)- c_{-i}v_i - c_{i}v_{-i}\right] + (K + c_{-i} - c_i)v_i + 0 \cdot v_{-i} \\ &=\left( \sum_{k} c_{-k}v_k \right)- \color{red}{c_{-i}v_i} - c_{i}v_{-i} + Kv_i + \color{red}{c_{-i}v_i} - c_i v_i\\ &=\left( \sum_{k} c_{-k}v_k \right) - c_{i}v_{-i} + Kv_i - c_i v_i\\ \end{align} That makes the $i$th entry of $Jv - Kv$ be \begin{align} (Jv - Kv)_i &=\left( \sum_{k} c_{-k}v_k \right) - c_{i}v_{-i} - c_i v_i\\ &=\left( \sum_{k} c_{-k}v_k \right) - c_{i}(v_{-i} + v_i) \end{align}

Now if $v$ is to be an eigenvector of eigenvalue $K$, then these numbers must all be zero. Picking $v_i = -v_{-i}$ ensures that the second part of each expression is zero; picking $v$ orthogonal to the vector $\bar{c}$ whose $i$th entry is $c_{-i}$ ensures that the first part is zero. So it's natural to ask "Can we do those two things simultaneously (without all entries being zero)?" (Even if we can't, there could still be an eigenvector, but this sure would be an easy way to find one!)

Let's pick all the $v_j$s to be zero except for $v_1, v_{-1}, v_2, v_{-2}$; we'll pick $v_{-1} = a = - v_1$ and $v_{-2} = b = -v_2$, so all the "second parts" will be zero. And now all we need is to find $a, b$, not both zero, satisfying
$$ a (c_{1} - c_{-1}) + b (c_2 - c_{-2}) = 0 $$

If either $c_{1} - c_{-1}$ or $c_{2} - c_{-2}$ is nonzero, we can choose $$ a = c_2 - c_{-2}\\ b = c_{-1} - c_1 $$ as a solution; if both are zero, we can pick $a = b = 1$ as a solution.

Conclusion: $K$ is indeed an eigenvalue of the jacobian. (Indeed, if in my solution above you pick indices other than $1$ and $2$, you find that $K$ is a multiple eigenvalue, at least when $N > 2$.)

Part 2: Conjecture 2 appears (to me) likely to be true.

Part 1 shows that if we define $\bar{c}$ to be the vector with $\bar{c}_i = c_{-i}$ (i.e., the vector $c$ "reversed"), and if we say that $v$ is "balanced" when $v_i + v_{-i} = 0$ for all $i$, then any balanced vector orthogonal to $\bar{c}$ is an eigenvector for $K = \sum_i c_i$.

Balanced vectors form a subspace of dimension $n$; those orthogonal to $\bar{c}$ form a subspace of dimension $2n -1$ (because $c$ is nonzero). The intersection of these has dimension at least $n - 1$ in $\Bbb R^{2n}$. Hence the eigenspace for $K$ has a subspace of dimension at least $n-1$ for which all vectors are balanced (i.e., for which conjecture 2 holds).

Playing around with then $N = 2$ case in matlab suggests to me that if the vector $c$ is "antibalanced" (i.e., if $c_i = c_{-i}$ for all $i$), then the eigenspace for $K$ has dimension $n$; otherwise it's got dimension $n-1$; I have not, however, proved this.

Now it's possible that the eigenspace for $K$ has dimension greater than $n$, in which case there might be lots of unbalanced solutions. But given that back in conjecture 1, we were wondering whether $K$ was even an eigenvalue (i.e., whether that space had dimension $> 0$), I suspect that this doesn't happen.

I also suspect that there might be some way to view all this as a problem involving directed graphs, where the answers are all obvious...but I don't see it.