# Homework Help: Linear Algebra: Geometric Interpretation of Self-Adjoint Operators

1. Apr 21, 2008

### smithg86

1. The problem statement, all variables and given/known data
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 $$| \lambda - \lambda ' | < \epsilon$$.

2. Relevant equations
n/a

3. The attempt at a solution
I thought this was similar to the statement that a function f(x) converges to a certain value(?)

2. Apr 21, 2008

### 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.

Last edited: Apr 21, 2008