why is it called an eigenvalue in this situation

53 Views Asked by At

In my book they study the following problem :

Let $r \in C^{\infty}([0,1])$, and let $\lambda$ be a real number, then we consider the following DE : $$y'' + (\lambda -r) y = 0$$

We denote $L_{\lambda}$ the vector space of solutions statifying $y(0) = y(1) = 0$, and if $L_{\lambda} \ne 0$ we say that $\lambda$ is an eigenvalue.

I am wondering why they call it an eigenvalue ? I mean for me an eigenvalue is when you have a matrix $A$ and a $\lambda$ such that : $Av = \lambda v$ for some $v \ne 0$. Yet here where is the matrix ?

Thank you !

1

There are 1 best solutions below

0
On BEST ANSWER

Generally speaking, an Eigenelement is an element such that a given transformation turns it to a multiple of itself, and the corresponding proportionality coefficient is an Eigenvalue.

In the case of a matrix applied to a vector,

$$Ax=\lambda x$$ associates an Eigenvector and an Eigenvalue. The transformation is described by a matrix.

In the case of your linear ODE,

$$\left(\frac{d^2}{dx^2}+r\right)y=\lambda y$$ associates an Eigenfunction and an Eigenvalue. The transformation is described by a differential operator.