In a paper that I'm reading, the author mentions something of this sort:-

...we arrive at an eigenvalue problem defined by the following matrix equation:

$ \left[ {\begin{array}{ccc} \sinh(\beta) & \sin(\gamma) \\ \beta \psi_\beta \sinh(\beta) & \gamma \psi_\gamma \sin(\gamma) \\ \end{array} } \right] \left[ {\begin{array}{cc} C_2\\ C_4\\ \end{array} } \right] =0 $

The eigenvalues are obtained by setting the determinant of the matrix to $0$ and then solving the characteristic equation. By solving the characteristic equation, one obtains $\sin(\gamma) = 0 \implies \gamma =n\pi $.

I admit I haven't had an introductory course in Linear Algebra yet, but I believe eigenvalues are $\lambda$ satisfying the matrix equation $Ax=\lambda x$, for some matrix $A$. But in the above case, the matrix equation is in the form $Ax=0$. Besides, obtaining $\sin(\gamma)=0$ is as simple as setting the determinant = $0$ (assuming that non-trivial solutions exist). How do eigenvalues and the characteristic equation come into the picture here?

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    $\begingroup$ I guess $\lambda=0$. $\endgroup$ Feb 14, 2014 at 12:51
  • $\begingroup$ Yes, I see that. But why even consider $\lambda$ here then? Why bring in the concept of eigenvalues here? How does it fit? $\endgroup$ Feb 14, 2014 at 12:57
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    $\begingroup$ What is it that you're trying to compute? $C_2$, $C_4$ or $\gamma$ and $\beta$? And also, are you asking here "Why not solve this as a system of equations"? $\endgroup$
    – frabala
    Feb 14, 2014 at 12:59
  • $\begingroup$ $\gamma$ and $\beta$. This is coming from a system of ODEs, whose general solution has $C_2$ and $C_4$ as two arbitrary constants. $\endgroup$ Feb 14, 2014 at 13:03
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    $\begingroup$ @JobinIdiculla I didn't think you were rude at all! Just curious, which is good. I think your question was perfectly natural. $\endgroup$ Feb 14, 2014 at 13:36

1 Answer 1


What I think is meant is something like this, with $B$ = your matrix, you can transform to an eigenvalue problem as follows $$\begin{align} B\mathbf{x} &= \mathbf{0} \\ (B-I)\mathbf{x} + \mathbf{x} &= \mathbf{0}\\ (B-I)\mathbf{x} &= -\mathbf{x},\;\;\;\;\;\;\text{now set }A=I-B\\ A\mathbf{x} &= \mathbf{x} \end{align}$$

The standard way to transform from an eigenvalue problem to the characteristic equation is the opposite of this. $$\begin{align} A\mathbf{x} &= \lambda\mathbf{x}\\ A\mathbf{x} - \lambda\mathbf{x} &= \mathbf{0}\\ (A-I\lambda)\mathbf{x} &= \mathbf{0} \end{align}$$ You already have an $= \mathbf{0}$ equation so I must admit to wanting to have more information than "we arrive at an eigenvalue problem".

  • $\begingroup$ Thank you for your answer. I think I slightly understand what was done, but I still don't get the whole picture. And I think you'd appreciate the context too. Could you have a look at pages 3-4 of the original paper here: 4shared.com/office/cIdBPJR5ce/… ? $\endgroup$ Feb 14, 2014 at 13:41
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    $\begingroup$ @JobinIdiculla Thankyou it's interesting to see the context. I've only skimmed the paper, but in many physical systems, you can get a solution to your governing equations as a set of independent eigenfunctions. The precise weighting of each one is determined by the ICs or BCs. So it seems like the motivation for talking about the problem as an eigenvalue problem might be to relate it to the physical system. Here it seems like each eigenfunction is a vibration, and each eigenvalue is the frequency of the corresponding vibration. $\endgroup$
    – TooTone
    Feb 14, 2014 at 14:25
  • $\begingroup$ Thank you so much for the help! I still have yet to understand it fully, but now I know what I'm looking for. $\endgroup$ Feb 14, 2014 at 17:22

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