Consider a matrix $V_t = I_{d\times d} + \sum_{i=1}^{t-1} x_i x^T_i$, where $x_i \in \mathbb{R}^d$ and $\|x_i\|_2 \leq 1$.
Is it possible to find upper bound (in terms of $d$ and $t$) for the ratio of the maximum and minimum eigenvalues of $V_t$ that is $\frac{\lambda_{max}(V_t)}{\lambda_{min}(V_t)}$?
I can prove $\det(V_t) \leq \left(1 + \frac{t}{d}\right)^d$ but not sure if it is useful for getting the bound.
We will restrict to the nontrivial case $d\ge 2$ (when $d=1$ then $\lambda_\min=\lambda_\max$). For $$V_t=I +\sum_{i=1}^{t-1}x_ix_i^T$$ we have $\lambda_{\min}\ge 1$ (because $V_t\ge I).$ On the other hand $$\lambda_{\max}=\max\{ \langle V_ty,y\rangle \,:\, \|y\|=1\}$$ For $\|y\|=1,$ by applying the Cauchy-Schwarz inequality, we obtain $$\langle V_ty,y\rangle=1+\sum_{i=1}^{t-1}\langle y,x_i\rangle ^2\le 1+\sum_{i=1}^{t-1}\|x_i\|^2\le t$$ Thus $${\lambda_\max\over \lambda_\min}\le t$$ The bound is optimal. Let $x_1=x_2=\ldots =x_{t-1}=e_1,$ where $e_1$ denotes the first vector of the standard basis. Then $$V_t=I+(t-1)e_1e_1^T$$ and $$\langle V_te_2,e_2\rangle =1,\qquad \langle V_te_1,e_1\rangle =t$$ Therefore $\displaystyle {\lambda_\max\over \lambda_\min}\ge t.$