Can't find this type of matrix in any index. It's the Jacobian of a non linear ODE system, and each row has only two row-specific values.

Question

I'm trying to understand this matrix's eigenvalues, but the matrix itself has a specific symmetry I had never seen before.

This matrix is $n \times n$, and the system has an integer $k \geq 1$ that uniquely characterizes it.

On each row in the matrix $J$, there are only two kinds of values:

• $J[i, i] = -\beta_i$, where $\beta_i$ is a real number.
• For $i \neq j$, we have $J[i,j] = \frac{\beta_i}{k}$.

The matrix, then, looks like this:

$$ J_{n\times n} = \begin{pmatrix} -\beta_1 & \frac{\beta_1}{k} & \frac{\beta_1}{k} & \cdots & \frac{\beta_1}{k} \\ \frac{\beta_2}{k} & -\beta_2 & \frac{\beta_2}{k} & \cdots & \frac{\beta_2}{k} \\ \frac{\beta_3}{k} & \frac{\beta_3}{k} & -\beta_3 & \cdots & \frac{\beta_3}{k} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{\beta_n}{k} & \frac{\beta_n}{k} & \frac{\beta_n}{k} & \cdots & -\beta_n \\ \end{pmatrix} $$

I know that there is no general expression for the eigenvalues of an $n \times n$ matrix when $n \geq 5$ (and in my case, $n = 90$). That being said, I was wondering if this kind of matrix symmetry has been studied and classified before (it must have been, right?). If that's the case, I was thinking that I could use that symmetry in order to characterize the eigenvalues somewhat. Maybe even derive an expression for them in terms of the $\beta_i$ values.

PS: I hope the matrix renders properly. I could not get the mathjax previewer to render the rows in separate lines. I hope I wrote it correctly, and the mathjax previewer is just messing with me :)

Answer 1

This matrix can be written as a sum $J = D + R$ where $D = \text{diag}(- \frac{k+1}{k} \beta_1, \dots)$ is a diagonal matrix and $R_{i, j} = \frac{\beta_i}{k}$ is a rank-$1$ matrix. We can write $R = uv^T$ where $u$ is the column vector with entries $\frac{1}{k}$ and $v^T$ is the row vector with entries $\beta_i$; this lets us compute the characteristic polynomial $\chi_J(\lambda)$ of $J$ using the matrix determinant lemma as

$$\begin{eqnarray*} \det(\lambda I - J) &=& \det((\lambda I - D) - uv^T) \\ &=& (1 - v^T (\lambda I - D)^{-1} u) \det (\lambda I - D). \end{eqnarray*}$$

$\lambda I - D$ is diagonal so the terms here are not hard to compute. We get

$$\det (\lambda I - D) = \prod_{i=1}^n (\lambda + \frac{k+1}{k} \beta_i), (\lambda I - D)^{-1} = \text{diag} \left( \frac{1}{\lambda + \frac{k+1}{k} \beta_1}, \dots \right)$$

so

$$v^T (\lambda I - D)^{-1} u = \sum_{i=1}^n \frac{\beta_i}{k (\lambda + \frac{k+1}{k} \beta_i)}$$

which gives that the characteristic polynomial $\chi_J(\lambda)$ is

$$\boxed{ \det(\lambda I - J) = \left( 1 - \sum_{i=1}^n \frac{\beta_i}{k (\lambda + \frac{k+1}{k} \beta_i)} \right) \prod_{i=1}^n \left( \lambda + \frac{k+1}{k} \beta_i \right) }.$$

If the $\beta_i$ are all real and have the same sign then the eigenvalues interlace with the values $\lambda_i = - \frac{k+1}{k} \beta_i$; you can see this by plugging in

$$\chi_J \left( - \frac{k+1}{k} \beta_i \right) = -\frac{\beta_i}{k} \prod_{j \neq i} \left( \frac{k+1}{k} \beta_j - \frac{k+1}{k} \beta_i \right).$$

In particular $n-1$ of the eigenvalues are real in this case so they're all real. As a sanity check, plugging in $\beta_1 = 1, \beta_2 = 2, \beta_3 = 3, k = 4$ and asking WolframAlpha for the eigenvalues gives eigenvalues of

$$-0.72 \dots, -1.96 \dots, -3.32 \dots $$

whereas the calculation above predicts that two of the eigenvalues lie between $- \frac{5}{4} = -1.25, -\frac{10}{4} = -2.5, -\frac{15}{4} = -3.75$, which checks out.

Answer 2

When multiplying a row by a constant, the determinant is multiplied by this constant. Therefore the determinant is equal to

$$ k^{-n} \prod_{j=1}^n \beta_i \underset{(*)}{\underbrace{\begin{vmatrix} -k & 1 & \dots & &1 \newline 1 & - k & \dots & &1 \newline &&\dots\newline 1 & & \dots & & -k \end{vmatrix}}} $$

Note that $(*)$ is the determinant of $J-(k+1)$, where $J$ is the all-ones matrix. The image of $J$ is one dimensional and therefore its null space is of dimension $n-1$, so that zero is an eigenvalue with geometric multiplicity $n-1$. Also, $\lambda = n$ is an eigenvalue. Thus, the eigenvalues of $J$ are $n,0,\dots,0$. This implies that the eigenvalues of $J-(k+1)$ are $n-k-1,-(k+1),\dots,-(k+1)$, and so its determinant $(*)$ is equal to their product, $(n-k-1)(-1)^{n-1}(k+1)^{n-1}$.

Bottom line, after doing the algebra:

$$ (-1)^{n-1}\frac{n-k-1}{k} (\frac{k+1}{k})^{n-1} \prod_{j=1}^n \beta_i.$$