Linear Algebra: Geometric Interpretation of Self-Adjoint Operators

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Homework Statement


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

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

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(?)
 

Answers and Replies

  • #2
Dick
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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.
 
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