I have a $d \times n$ matrix $X$ consisting of $n$ vectors $x_i \in \mathbb{R}^d$. That is, $X = [x_1 x_2 \ldots x_{n-1} x_n]$. I also know that $\|x_i\|_2 \leq 1$. Let's say I have another matrix $X'$, which contains only one different entry than $X$. Say, $X' = [x_1 x_2 \ldots x_{n-1} x'_n]$. Let us define $A = XX^\top = \sum_{i=1}^n x_i x_i^\top = \sum_{i=1}^{n-1} x_i x_i^\top + x_n x_n^\top$. Similarly, $A' = \sum_{i=1}^{n-1} x_i x_i^\top + x_n' x_n'^\top$. Now, I need to find an upper-bound for $\text{tr}(A-A')$. We have, $\text{tr}(A-A') = \text{tr}(x_n x_n^\top - x_n' x_n'^\top) = \|x_n\|_2^2 - \|x_n'\|_2^2$. This is upper-bounded by $1$ because each sample $x_i$ has $\mathcal{L}_2$-norm at most $1$. But this bound is very loose in practice. Can anyone please suggest me a stronger bound?
Please note that $x_i$'s are given to me, that is, they are not random and I do not know their distribution.