# How is this an eigenvalue problem?

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?

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

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".