Let $\left\{u\_i\right\}\_{i=1}^\infty$ be a sequence of real vectors (i.e. $u\_i\in R^n, i=1,2,... $) and $m$ an integer large enough such that $\sum\_{i=1}^m u\_i u\_i^T$ is a positive definite matrix. Define:


$$K\_m:=\left(\sum\_{i=1}^m u\_i u\_i^T\right)^{-1}u\_m$$


Show that $\left\| K\_m\right\| $ tends to zero as $m\longrightarrow \infty$.


If $u\_i$s are scalar, i.e. $n=1$, the proof seems trivial (not *that* trivial, but easy to get). I have not been able to find a counter-example for the cases where $u\_i$s are vectors (and believe me, I have tried hard). So I think this is very likely to be true in general.



Needing proof of convergence for a sequence
Let me first discuss the special case where $|u\_n|$ is bounded. Let's denote our matrices by $A\_n=\sum\_{j=1}^n u\_ju\_j^\*$.
What we're trying to show can only fail if there are infinitely many $u\_n=c\_nv\_n + w\_n$ that have a coefficient $|c\_n|\ge\delta >0$ when expanded in terms of the eigenvectors of $A\_n$, such that the corresponding eigenvalue $\lambda\_{n}$ stays bounded.


By compactness, these eigenvectors $v\_n$ converge to a vector $v$ on a subsequence. Then, however, $\langle v\_n, A\_n v\_n\rangle$ will become large because each of the many summands $u\_ju\_j^\*$ of $A\_n$ with a $u\_j$ that is as above (with a component $\ge\delta$ in almost the direction of $v\_n$) makes a contribution $\gtrsim \delta^4$ to this quadratic form. So $\lambda\_n$ can not stay bounded.


In general, if $|u|$ is large, then consider (writing $Q=uu^\*$)
$$
|A^{-1/2}u|^2 = |(Q+B)^{-1/2}u|^2 =\langle u, (Q+B)^{-1} u\rangle ;
$$
this is bounded because $-1/z$ is operator monotone, so $B$ in the quadratic form can be essentially dropped (more precisely, replace it by $\epsilon B$).


In other words, $x\_n=A\_n^{-1/2}u\_n$ is always bounded, and we still find ourselves in the same scenario as above: to prevent $A\_n^{-1/2}x\_n$ from going to zero, $x\_n$ and thus also $u\_n$ must have a component bounded away from zero belonging to a bounded eigenvalue of $A\_n$.


