## Eigenvalue

"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.
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 Recognitions: Homework Help Science Advisor Suppose $|\lambda _i - \lambda| \geq \epsilon$ for all i, then: $$\epsilon^2 > |(\lambda_1-\lambda)x_1|^2+...+|(\lambda_n-\lambda)x_n|^2 = \sum _{k=1} ^n |\lambda _k - \lambda|^2|x_k|^2 \geq \sum \epsilon ^2|x_k|^2 = \epsilon ^2\sum |x_k|^2 = \epsilon ^2 ||v|| = \epsilon ^2$$