Suppose for some matrix $A\in\mathbb{R}^{n\times n}$, and some vectors $x,y\in\mathbb{R}^n$, $y=Ax$.
Under what conditions (ideally on $A$) can one construct a matrix $B\in\mathbb{R}^{n\times n}$ such that it is also true that $y=Bx$, but where the maximum singular value of $B$ is less or equal to $1$?
This is obviously trivial for $A$ with maximum singular value less than or equal to $1$, or when $\|y\|\le\|x\|$. Could it hold more generally?
I started tackling this along the following lines:
Let $B=UDV^*$ be the singular value decomposition of $B$. Then for all $i\in\{1,\dots,n\}$, $D_{ii}=\frac{U_{\bullet i}^*y}{V_{\bullet i}^*x}$, providing the denominator is non-zero. Thus, we need to find unitary matrices $U$ and $V$ such that (WLOG) $U^*y\ge 0$, $V^*x\gg 0$ and $U^*y\le V^*x$.
Since $V_{\bullet i}^*x\le\|x\|$, for all $i\in\{1,\dots,n\}$, a necessary condition for a solution is that $U_{\bullet i}^*y\le\|x\|$, for all $i\in\{1,\dots,n\}$. Hence, if $\theta_i$ is the angle between $U_{\bullet i}$ and $y$, we require that $\cos{\theta_i}\le\frac{\|x\|}{\|y\|}$, for all $i\in\{1,\dots,n\}$. This it turn implies that it must at least be the case that $\sqrt{n}\ge\frac{\|y\|}{\|x\|}$.
If deriving such a $B$ is possible, I have a few follow on questions. Can we make the inequality strict? Can we practically choose $B$ to minimise its distance to $A$ according to some norm, while satisfying these criteria? (I know there's always the option of setting up a numerical optimisation problem.)