Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
1  
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
add comment

2 Answers

up vote 3 down vote accepted

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 $Ax=0, x \neq 0$.

share|improve this answer
    
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
add comment

Find a vector $v$ such that :

1) $v$ is not zero ;

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

share|improve this answer
1  
$A$ 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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