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.