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I'm having difficulties with this homework problem:

Suppose $T$ is a self adjoint linear operator on a vector space $V$. Let $\lambda \in \mathbb{F}$ (where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$; naturally $V$ is a vector space over $\mathbb{F}$) and $\epsilon > 0$. Prove that if there exists $v \in V$ such that $||v|| = 1$ and $||Tv-\lambda v|| < \epsilon$, then $T$ has an eigenvalue $\lambda'$ such that $|\lambda - \lambda'|<\epsilon$.

I thought I had proven it, but I assumed that $v$ is an eigenvector of $T$ with eigenvalue $\lambda'$ (in fact then it is quite simple, since it turns out $||Tv-\lambda v|| = |\lambda - \lambda'|$). So here is my question: can we assume without loss of generality that $v$ is an eigenvector with eigenvalue $\lambda'$? I suspect the answer is no, so if not I'd also appreciate a push in the right direction.

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No, you cannot just assume that $v$ is an eigenvector, because you are not told that the inequality holds for all $v\in V$ of norm $1$, just for a particular one. Maybe it holds for one that is not an eigenvector at all? – Arturo Magidin Nov 17 '11 at 17:55
up vote 1 down vote accepted

The following is for finite dimensional vector spaces.

Replacing $T$ with $T-\lambda I$ we may assume that $\lambda=0$, and that the conditions are that $\lVert v\rVert=1$, $\lVert Tv\rVert \lt \epsilon$, and we want to prove there is an eigenvalue $\lambda'$ with $|\lambda'|\lt \epsilon$.

Since $T$ is self-adjoint, it is orthogonally diagonalizable. Let $v_1,\ldots,v_n$ be an orthonormal basis of eigenvectors, with corresponding eigenvalues $\lambda_1,\ldots,\lambda_n$. We may assume that $|\lambda_1|\leq\cdots\leq |\lambda _n|$.

Write $v$ as a linear combination of $v_1,\ldots,v_n$, $$v = \alpha_1v_1+\cdots + \alpha_nv_n.$$ Since $\lVert v\rVert = 1$, then $|\alpha_1|^2+\cdots+|\alpha_n|^2 = 1$.

Now, $$Tv = \alpha_1\lambda_1v_1+\cdots + \alpha_n\lambda_nv_n,$$ hence $$\lVert Tv\rVert^2 = |\alpha_1\lambda_1|^2 + \cdots + |\alpha_n\lambda_n|^2 \geq (|\alpha_1|^2+\cdots+|\alpha_n|^2)|\lambda_1|^2.$$

Can you take it from here?

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This is ok if the vector space is finite-dimensional. For a self-adjoint linear operator on an infinite-dimensional Hilbert space, there may be no eigenvalues at all. For the result to be true you'd have to replace "eigenvalue" by "member of the spectrum". – Robert Israel Nov 17 '11 at 18:24
So $\epsilon^2 > ||Tv||^2 \ge (|\alpha_1|^2+\cdots+|\alpha_n|^2)|\lambda_1|^2 \ge |\lambda_1|^2$, and choosing $\lambda'=\lambda_1$ we see that $|\lambda'| < \epsilon$. Thanks! By the way, we can assume that $V$ is finite dimensional ... but I'd be interested to hear the argument if it's not. – smackcrane Nov 17 '11 at 18:31
@smackcrane: Sorry; it's been too long since I played around with spectra in Hilbert spaces. Perhaps Robert Israel can post an answer in that case. – Arturo Magidin Nov 17 '11 at 20:42

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