Linear Algebra: Geometric Interpretation of Self-Adjoint Operators

[smithg86]

Homework Statement

I'm not interested in the proof of this statement, just its geometric meaning (if it has one):

Suppose T \in L(V) is self-adjoint, \lambda \in F, and \epsilon > 0. If there exists v \in V such that ||v|| = 1 and || Tv - \lambda v || < \epsilon, then T has an eigenvalue \lambda ' such that | \lambda - \lambda ' | < \epsilon.

Homework Equations

n/a

The Attempt at a Solution

I thought this was similar to the statement that a function f(x) converges to a certain value(?)

 

[Dick]

Eigenvectors are vectors v, such that T(v) is a multiple of v and the the eigenvalues are those constant multiples. This says that if you can find a unit vector v, such that T(v) is 'almost' a multiple of itself (lambda*v), then lambda is 'almost' an eigenvalue.