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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.

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    $\begingroup$ 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

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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:

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

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  • $\begingroup$ 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

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