Bounding $x^\top Ay/x^\top y$

Question

The Rayleigh-Ritz theorem maximises $x^\top A x/x^\top x$ and expresses the result in terms of eigenvalues of $A$. Are there theorems which study $$x^\top Ay/x^\top y,\ x^\top y\ne 0$$ for two given vectors $x$ and $y$ and bound the term in terms of the spectrum of $A$?

Answer 1

Both $$A=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, B=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$

have eigenvalues 1

But for the first one, your expression is always 1, while for the second one, take $x=\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, and $y_\epsilon=\begin{pmatrix} \epsilon \\ 1 \end{pmatrix}$

then $x^\bot y_\epsilon = \epsilon$, and $x^\bot By_\epsilon = 1+\epsilon$

So the quotient is unbounded when $\epsilon \to 0$

Answer 2

If $A$ is not a scalar matrix, then there exists a $x\in F^n$ such that $x^\top A \nparallel x^\top$, fix such a $x$, then there exists a $y\in F^n$ such that $x^\top Ay=0$ while $x^\top y\neq 0$. Thus the infimum of $\Vert x^\top Ay/x^\top y\Vert$ is always $0$ unless $A$ is a scalar matrix. If $A$ is invertible and not a scalar matrix, then the supreumum of $\Vert x^\top Ay/x^\top y\Vert$ equals to the supreumum of $\frac 1{\Vert x^\top A^{-1}y/x^\top y\Vert}$ equals to $0$. So I think such a study is not so interesting.