TITLE: Needing proof of convergence for a sequence
QUESTION [14 upvotes]: 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.

REPLY [3 votes]: 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 $A_n$.