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Let $V$ be an inner-product vector space (real, of very large but finite dimension, if you wish). Let $S:V\to V$ be a symmetric linear operator. Let $v_1,\dotsc,v_k\in V$ be a collection of vectors of norm $1$, orthogonal to each other. (Here $k$ may be thought of as larger than a constant, but possibly much smaller than $\dim V$.) Assume that

$(S v_i,v_i)>\delta$ for every $1\leq i\leq k$

and $|S v|_2\leq |v|_2$ for every $v\in V$.

Question: Can we conclude that the space W spanned by all eigenspaces with eigenvalue $\geq \delta/2$ (say; $\delta^2$ or $\delta^{100}/100$ would also do) has large dimension?

By "large" I mean something in the scale of $k$, $\sqrt{k}$, $\delta k$ or thereabouts. It is easy to give a very weak bound ($\gg \log(O(\delta^2 k))$; see below).

Question, version 2: The same question, but with the added condition $(S v_i, v_j)=0$ for all distinct $1\leq i,j\leq k$.


Sketch of proof of the weak bound: Suppose $W$ had dimension $d=o(\log k)$. By pigeonhole, there is narrow cone in $W$ such that the projections $w_i$ to $W$ of $\geq k/6^d > \sqrt{k}$ vectors $v_i$ lie in that cap. Any two vectors in that cap have inner product $\geq 1/2$ times the product of their norms. Hence $\langle w_i,w_j\rangle \geq |w_1|_2 |w_2|_2/2 \geq |\delta|^2/8$. Then the projections $u_i$ of those same vectors $v_i$ to the orthogonal complement $U$ of $W$ satisfy $\langle u_i,u_j\rangle \leq -|\delta|^2/8$ for all $i$, $j$ distinct. By a standard argument, this condition can be satisfied by at most $O(|\delta|^{-2})$ vectors.

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Write $P$ for the projection onto the $v_j$, and let $N$ denote the number of eigenvalues $\ge \delta/2$ of the matrix $PSP$. Then $$ k\delta \le\textrm{tr}\: PSP \le N + \frac{(k-N)\delta}{2} , $$ so $N\ge \delta k/3$, say, for small $\delta$. By the min-max principle, the same inequality holds for the original matrix.

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  • $\begingroup$ I addressed this last step more explicitly in this answer of mine (which I'm fairly confident is correct, even though the OP recently felt compelled to unaccept it): mathoverflow.net/questions/248198/… $\endgroup$ – Christian Remling Feb 17 '17 at 4:21
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Actually, isn't question 2 brutally trivial? The following argument seems to show that the space $W$ spanned by the eigenspaces with eigenvalue $\geq \delta$ has dimension $\geq k$.

Suppose this weren't the case. Then there would be a non-trivial linear combination $\sum_i a_i v_i$ orthogonal to $W$. Thus, on the one hand,

$$\left\langle S\left(\sum_i a_i v_i\right), \sum_i a_i v_i\right\rangle < \delta \left|\sum_i a_i v_i\right|^2 = \delta \sum_i |a_i|^2,$$

but, on the other hand,

$$\begin{aligned}\left\langle S\left(\sum_i a_i v_i\right), \sum_i a_i v_i\right\rangle &= \sum_{i,j} \overline{a_i} a_{j} \langle S v_i, v_{j}\rangle\\ &= \sum_i |a_i|^2 \langle S v_i, v_i\rangle \geq \delta \sum |a_i|^2.\end{aligned}$$

Contradiction.

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