Eigenvalues of an operator in an inner product space

[Treadstone 71]

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

||T(v)-\lambda v||<\epsilon.

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

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

||(\lambda_1-\lambda)x_1e_1+...+(\lambda_n-\lambda)x_1e_1||<\epsilon

Since the basis is orthonormal, it follow that

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

At this point I am unable to deduce the conclusion.

 

[AKG]

Suppose |\lambda _i - \lambda| \geq \epsilon for all i, then:

\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