# 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.

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 :)

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.

• Thank you so much! Later during the week, I noticed I had computed the matrix slightly wrong; however, the row pattern ended up being the same. Thanks to your answer, I was able to follow the same steps and found an expression for the characteristic polynomial. I think that helped me to understand this ODE system a lot better too. Again, thank you so much! Commented Aug 19 at 17:45

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:

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

• That only computes the determinant, not the eigenvalues. Commented Aug 12 at 18:47