I have a sort of reverse problem, and I'm not sure if there is a simple solution.

I have a tridiagonal Hermitian matrix: $$ A = \begin{bmatrix} 0 & a_1 & 0 & 0 & 0 \\ a_1 & 0 & a_2 & 0 & 0 \\ 0 & a_2 & 0 & a_3 & 0 \\ 0 & 0 & a_3 & 0 & a_n \\ 0 & 0 & 0 & a_n & 0 \\ \end{bmatrix} $$ where $a_n$ are unknown; however, I know $A$'s eigenvalues: $$ \lambda = \{\lambda_1, \lambda_2, \lambda_3, \lambda_n\}. $$

I want to find the $a_n$'s that would give me the known eigenvalues that I have. So I'm after the matrix that gives me these eigenvalues. I also know the order of the eigenvalues, so I that, say, the first eigenvalue belongs to the largest eigenvector, and the second to the second largest eigenvector. I don't know the eigenvectors though, and I don't really care what they are.

I've looked into closed-form equations to calculate the characteristic polynomials for a matrix of the form of $A$; however, I'm not sure how easy it is to go from these to $a_n$'s.

Any advice would be much appreciated.

  • 1
    $\begingroup$ How do you determine the "largest eigenvector"? As far as I know the eigenvector magnitude is any positive number. $\endgroup$
    – nicoguaro
    Apr 7, 2016 at 17:12
  • $\begingroup$ Also, it seems that some eigenvalues are negative and some positive, since the trace of the matrix is zero. $\endgroup$
    – nicoguaro
    Apr 7, 2016 at 17:30
  • $\begingroup$ If you know all the eigenvalues (and they are all unique), you can easily compute the eigenvectors and formulate them into a matrix $Q$ consisting of the normalized eigenvectors. Then, $A=Q\Lambda Q^{-1}$, where $\Lambda$ is the diagonal matrix of eigenvalues. $\endgroup$
    – Paul
    Apr 7, 2016 at 17:36
  • $\begingroup$ You're looking at a Jacobi matrix. These matrices have a number of interesting properties, however, I'm not sure they are of much help for the problem at hand. What's certain is that the direct problem (find $\lambda$ given the $a_i$'s) has no closed-form solution. You'll find more details in this older math.stackexchange post. $\endgroup$
    – GoHokies
    Apr 7, 2016 at 19:10
  • $\begingroup$ There are a number of papers discussing inverse problems for Jacobi operators (i.e., recovery of the Jacobi matrix, given its eigenvalues). Here is one to get the ball rolling... $\endgroup$
    – GoHokies
    Apr 7, 2016 at 19:32

1 Answer 1


As a first thought, you could formulate the problem by creating the following set of nonlinear equations:

$$g_i(\textbf{a}) = det(A(\textbf{a})-\lambda_i I)=0 \;\;\;\;\;\forall i \in [1,n]$$

Then you could try solving it through some root-finding or optimization approach. Note that there's a recursive formula for Tridiagonal determinants that should reduce the above equations to the following:

$$g_i(\textbf{a}) = f_{n}^{(i)}=0 \;\;\;\;\;\forall i \in [1,n]$$ $$where \;\;\;\;\;f_{j}^{(i)}=-\lambda_i f^{(i)}_{j-1} - a_{j-1}^2 f^{(i)}_{j-1}, f^{(i)}_0 = 1, f^{(i)}_{-1} = 0 \;\;\;\;\;\forall i \in [1,n]$$

  • $\begingroup$ I was actually thinking along these lines. Do you know if there's could be any issue with making sure that eigenvalues belong to certain eigenvectors? I don't know how to incorporate that info into this, or if it even makes a difference. $\endgroup$
    – Jean-Luc
    Apr 7, 2016 at 22:34
  • $\begingroup$ Actually thinking about it somewhere, I shouldn't need to care about which eigenvectors the eigenvalues correspond to; it should work out. $\endgroup$
    – Jean-Luc
    Apr 7, 2016 at 23:05
  • $\begingroup$ Given the eigenvectors don't come up in the formulation above, I suspect it shouldn't matter. So, I too agree it should just work out. $\endgroup$
    – spektr
    Apr 7, 2016 at 23:48
  • $\begingroup$ Yep cool! I'm implenting it now! Thanks. By the way, just for completeness, I think on the last line in your comment/equation, there should be a negative sign in front of $\lambda_i$. $\endgroup$
    – Jean-Luc
    Apr 9, 2016 at 12:02
  • $\begingroup$ Oh you're right! I will change that! $\endgroup$
    – spektr
    Apr 9, 2016 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.