Consider the matrix $$ M := \left[ \begin{matrix} 1 & \epsilon & 2\epsilon \\ \epsilon & 1 & 2\epsilon \\ -2\epsilon & - 2\epsilon & 2 \end{matrix} \right] $$ for some small parameter $0 < \epsilon \ll 1$. My question is how does one solve for the eigenvalues $\lambda$ of this matrix as a series in $\epsilon$, given that the zeroth-order (with $\epsilon = 0$) eigenvalues have degeneracy?

I am more interested in the procedure than anything else here. The eigenvalues are exactly given by $$ \lambda \in \left\{ 1 - \epsilon, \frac{3 + \epsilon - \sqrt{ 1 - 2 \epsilon - 31 \epsilon^2}}{2} , \frac{3 + \epsilon + \sqrt{ 1 - 2 \epsilon - 31 \epsilon^2}}{2} \right\} \ , $$ which when expanded for $0 < \epsilon \ll 1$ gives respectively $$ \lambda \in \left\{ 1 - \epsilon, 1 + \epsilon + 8 \epsilon^2 + \mathcal{O}(\epsilon^3) , 2 - 8\epsilon^2 + \mathcal{O}(\epsilon^3) \right\} \ . $$ How would one solve for these eigenvalues as a series if it was not possible to solve for the exact eigenvalues?

My attempt: Normally what I would do would be to assume I have a series $$ \lambda = \lambda_0 + \epsilon \lambda_1 + \epsilon^2 \lambda_2 + \ldots $$ and insert this series into the characteristic equation for the matrix which is here $$ \lambda^3 - 4 \lambda^2 + (5 + 7 \epsilon^2) \lambda -2 - 6 \epsilon^2 + 8 \epsilon^3 = 0 \ . $$ However this method yields nonsense because of the degeneracy of the first two eigenvalues (both being 1) in the limit that $\epsilon = 0$. To describe what happens: the characteristic equation tells me that $\lambda_0^3 - 4 \lambda_0^2 + 5 \lambda_0 - 2 = 0$ for $\epsilon =0$, which gives me that $\lambda_0 \in \{1,1,2\}$. However, at next order in $\epsilon$ in the characteristic equation tells me $5 \lambda_1 - 8 \lambda_0\lambda_1 + 3 \lambda_0^2 \lambda_1 = 0$, which when evaluated for the degenerate $\lambda_0 = 1$ just says that $0=0$ and so gives no information at all, and so I cannot proceed to find $\lambda_1$.

  • $\begingroup$ I would expect the (2,1) entry to have a minus sign $\endgroup$
    – Will Jagy
    Sep 1, 2021 at 2:05
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    $\begingroup$ +1 Interesting question. My initial thought is that it's not going to be plausible, at least outside the symmetric case, as such a method should fall apart when the eigenvalues become complex. It's difficult to see how a nice method would account for things like radius of convergence of the series as well. As always, I could be wrong though. $\endgroup$ Sep 1, 2021 at 2:07
  • $\begingroup$ @WillJagy I purposefully cooked up that entry have that sign, because the exact eigenvalues ended up being simpler in that case. I agree that the matrix would like nicer with that sign though $\endgroup$ Sep 1, 2021 at 2:29
  • $\begingroup$ @TheoBendit Okay thanks for your thoughts. I agree that the complicating factor here is the fact that this matrix is non-symmetric (in quantum mechanics, there is a formalism for dealing with degenerate perturbation theory analogous to the above, but only for the symmetric/Hermitian case) $\endgroup$ Sep 1, 2021 at 2:31
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    $\begingroup$ One other point to consider is that the matrix always possesses the eigenvector $(1,-1,0)$ for eigenvalue $1-\epsilon$. So this eigenvalue is never problematic. That suggests projecting the matrix onto the orthogonal complement of this subspace. Doing so gives the matrix $$\begin{pmatrix} 1+\epsilon & 2\epsilon\\ -4\epsilon & 2\end{pmatrix},$$ which retains the two problematic eigenvalues. $\endgroup$ Sep 1, 2021 at 4:13

1 Answer 1


This is really a question about the polynomial, not about the matrix. Consider a simple example $$(x-1)^2-a \epsilon^2 = 0$$ and expand: $x=1+x_1 \epsilon+ x_2 \epsilon^2 \cdots $. The first order in $\epsilon$ is indeed $0=0$. This does not mean you have "nonsense", it just means that $x_1$ is determined by the next order: $$x_1 ^2 \epsilon^2 -a \epsilon^2 =0$$ Similarly, for your more complicated example: $$\lambda^3 - 4 \lambda^2 + (5 + 7 \epsilon^2) \lambda -2 - 6 \epsilon^2 + 8 \epsilon^3 = 0 \ $$ expanding to second order around $\lambda = 1 + \lambda_1 \epsilon +\cdots$ will give $1-(\lambda_1)^2 = 0$ as expected.

The reason is the double zero: there is a parabola $y=x^2$ that just touches the $x$ axis. An order $\epsilon$ perturbation in the constant $x^2 -\epsilon$ yields a $O(\sqrt \epsilon)$ in the root. Luckily for you, the perturbation is $x^2- \epsilon^2$, so you get an $O(\epsilon)$ change in the root position.


As an example of how things could have been even worse, consider the matrix $$\pmatrix{1&\epsilon\\a &1} $$ with the polynomial $$(1-\lambda)^2-a\epsilon$$ the lowest order term in the expansion is $1\pm\sqrt a \epsilon^{1\over2}$, so that the correction is a half order!


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