Bounding the difference of rank-1 matrices

Question

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.

Answer 1

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 $\|x\|_2^2 - \|y\|_2^2$.