Eigenvalues of an operator in an inner product space

[Treadstone 71]

"Suppose [tex]V[/tex] is a (real or complex) inner product space, and that [tex]T:V\rightarrow V[/tex] is self adjoint. Suppose that there is a vector [tex]v[/tex] with [tex]||v||=1[/tex], a scalar [tex]\lambda\in F[/tex] and a real [tex]\epsilon >0[/tex] such that

[tex]||T(v)-\lambda v||<\epsilon[/tex].

Show that T has an eigenvalue [tex]\lambda '[/tex] such that [tex]|\lambda -\lambda '| < \epsilon[/tex]."

Since T is self adjoint, there exists an orthonormal basis [tex](e_1,...,e_n)[/tex], with corresponding eigenvalues [tex]\lambda_1,...,\lambda_n[/tex]. Suppose [tex]v=x_1e_1+...+x_ne_n[/tex] for some [tex]x_1,...,x_n\in F[/tex]. Then,

[tex]||(\lambda_1-\lambda)x_1e_1+...+(\lambda_n-\lambda)x_1e_1||<\epsilon[/tex]

Since the basis is orthonormal, it follow that

[tex]|(\lambda_1-\lambda)x_1|^2+...+|(\lambda_n-\lambda)x_n|^2<\epsilon^2[/tex].

At this point I am unable to deduce the conclusion.

 

[AKG]

Suppose [itex]|\lambda _i - \lambda| \geq \epsilon[/itex] for all i, then:

[tex]\epsilon^2 > |(\lambda_1-\lambda)x_1|^2+...+|(\lambda_n-\lambda)x_n|^2 = \sum _{k=1} ^n |\lambda _k - \lambda|^2|x_k|^2 \geq \sum \epsilon ^2|x_k|^2 = \epsilon ^2\sum |x_k|^2 = \epsilon ^2 ||v|| = \epsilon ^2[/tex]