# Eigenvalue of some Sturm–Liouville problem

I have one simple question. How I suppose to show that $\lambda =0$ is an eigenvalue of some problem. Does it mean that I must have non-trivial solution for $\lambda=0$?

Thanks!

UPD:I mean by problem some Sturm–Liouville equation...

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If your (hopefully) matrix is low dimensional, you may try proving determinant is zero. – Tapu Feb 4 '13 at 17:08
@ Tapu I mean by problem some Sturm–Liouville equation – Panka Feb 4 '13 at 20:00

Suppose that $A$ is a linear operator of some vector space. Then $\lambda$ is an eigenvalue if and only if there exists non-zero vector $x$ such that $Ax= \lambda x$. In the case of $\lambda=0$, we must have $Ax=0, x \neq 0$.

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I mean by problem some Sturm–Liouville equation, in this case i need to have non-trivial solution? Thanks – Panka Feb 4 '13 at 20:02
Yes, non-trivial, that is, non-zero. – Manos Feb 4 '13 at 21:07
@ Manos Thank you! – Panka Feb 4 '13 at 21:25

Find a vector $v$ such that :

1) $v$ is not zero ;

2) $Av = 0$, where $A$ is the matrix in your problem.

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$A$ does not have to be a matrix, could be any linear operator -- as in Manos's answer – gt6989b Feb 4 '13 at 17:08
Sure, but I am keeping it simple for Panka. – Damien L Feb 4 '13 at 17:10

Yes, you must have a non-trivial solution for $\lambda=0$. For example, the SL problem $y^{\prime\prime}+\lambda y=0$, with boundary conditions $y^{\prime}(0)=0$, $y^{\prime}(1)=0$ has $\lambda=0$ for an eigenvalue with $y(x)=c$, where $c \neq 0$ is a constant for an eigenfunction.

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