Let $x, y$ be vectors in $\mathbb{R}^n$ and $A \in \mathbb{R}^{n \times n}$ is a PSD matrix.
I would like to bound
$$|y^TAyx^TAx - (x^TAy)^2| \leq ?$$
for a fixed $A, x$ with a varying $y$. For example, if we let $||y||\_2 \leq L$, can we give a bound in terms of $A$, $x$, and $L$?


Any pointers even just related are appreciated, thanks!



Upper bound for $|y^TAyx^TAx - (x^TAy)^2|$ where A is PSD?
By the Cauchy--Schwarz inequality, $y^TAy\,x^TAx\ge(x^TAy)^2$. So,
$$|y^TAyx^TAx-(x^TAy)^2|=y^TAyx^TAx-(x^TAy)^2
=y^T[(x^TAx)A-Axx^TA]y.$$
So, for real $L\ge0$, the best upper bound on $|y^TAyx^TAx-(x^TAy)^2|$ given $\|y\|\_2\le L$ is
$$\max\_{\|y\|\_2\le L}y^T[(x^TAx)A-Axx^TA]y=L^2\|(x^TAx)A-Axx^TA\|
,$$
where $\|(x^TAx)A-Axx^TA\|$ is the largest eigenvalue of the positive-semidefinite matrix $(x^TAx)A-Axx^TA$.


By the spectral decomposition, the calculation of $\|(x^TAx)A-Axx^TA\|$ reduces to the case when $A$ is diagonal. A trivial upper bound on $\|(x^TAx)A-Axx^TA\|$ is $(x^TAx)\|A\|$; this bound will work well when $n$ is not small, the eigenvalues of $A$ do not differ much from one another, and the coordinates of $x$ in an eigenbasis of $A$ do not differ much from one another in absolute value.


