2

Let $(x_i)_{1 \leq i \leq n}$ be vectors of $\mathbb{R}^d$. Assume that $\sum_{i}x_ix_i^T$ is invertible and let $A=(\sum_{i}x_ix_i^T)^{-1}$. Let $v$ be a non-zero vector of $\mathbb{R}^d$.

A lower bound of $v^TAv$ can be found by noticing that $\sum_{i} x_ix_i^T \preceq \left(\sum_{i} x_i^Tx_i\right)I_d$, which yields $A \succeq \left(\sum_{i} x_i^Tx_i\right)^{-1}I_d$ and $$v^TAv \geq \left(\sum_{i} x_i^Tx_i\right)^{-1}v^Tv.$$ I would like to find an upper bound of $v^TAv$ that has a similar shape as the lower bound (I'd like it to be quite explicit in the vectors $x_i$ and in $v$)

I have noticed that if we denote by $\lambda_{min}(M)$ (resp. $\lambda_{max}(M)$) the smallest (resp. largest) eigenvalue of a matrix $M$, we have $$v^TAv \leq \lambda_{max}(A) v^Tv = \frac{1}{\lambda_{min}\left(\sum_{i}x_ix_i^T\right)}v^Tv.$$

However, this is not very satisfactory to me because it is not "explicit enough" in terms of the vectors $x_i$. Perhaps there could be a non-trivial lower bound of the aforementioned smallest eigenvalue ?

Skywear
  • 383
  • 1
    Please, use MathJax also in the title. – jjagmath Jun 23 '23 at 16:17
  • 1
    There is no upper bound which is $|v|^2$ times a function only of $|x_1|,\cdots,|x_n|$ like your lower bound is. To see this, consider $x_1=(1,0)$ and $x_2=(\cos\theta,\sin\theta)$ in 2D as $\theta\to0$. – coiso Jun 23 '23 at 16:34
  • @elemelons thanks, this is insightful. If you write it as an answer i'll accept it. – Skywear Jun 24 '23 at 08:18

0 Answers0