# Bounding the difference of rank-1 matrices

Let $$\|\cdot\|$$ denote the Frobenius norm and $$x,y \in \mathbb{R}^n$$

I need a bound of the form

$$\|x \cdot x^\top - y \cdot y^\top \| \leq C \|x-y\|_2 \quad (*),$$

where $$C>0$$ does not depend on $$x$$ or $$y$$.

This seems to work when $$\|x\|_2=\|y\|_2 = 1$$. (Then I get $$C=2$$ by basic linear algebra).

For general $$x,y$$ this seems not to be possible, which can already be seen for $$n=1$$.

I wonder what is known regarding more advanced bounds on the LHS (like eigenvalues of sums of rank-1 matrices?), that comes as close as possible to the RHS of (*).

Thanks very much for any help or suggestion on this.

For the Frobenius norm, we have $$\|xy^T\|_F^2 = \sum_{i,j}(x_i y_j)^2 = \|x\|_2^2 \|y\|_2^2.$$ This implies $$\|xx^T-yy^T\|_F= \|(x-y)x^T+y(x-y)^T\|_F \le \|(x-y)x^T\|_F+\|y(x-y)^T\|_F \le (\|x\|_2 + \|y\|_2)(\|x-y\|_2).$$ The constant cannot be independent of $$x,y$$, because the mapping $$x\mapsto xx^T$$ is 'quadratic' in $$x$$.
Taking the difference of the non-zero eigenvalues of the rank-one matrices does not help, as this difference is $$\|x\|_2^2 - \|y\|_2^2$$.