Neighborhood of an orthogonal matrix

Question

Let $A\in O(n)$ be an orthogonal matrix and let $\vec{a}_1,\dots,\vec{a}_n$ be its rows. For a vector $\vec{v}=[v_1,\dots,v_n]$, let $\max(\vec{v})=\max\{|v_1|,\dots,|v_n|\}$. Prove or disprove that if $\max(\vec{a}_i)<1$ for every $i$, then for every small neighborhood of $A$ there exists $B\in O(n)$ such that $\max(\vec{b}_i)>\max(\vec{a}_i)$ for every $i$.

For me the general approach to this kind of problem is to form a map $f:O(n)\to\mathbb{R}^n$ where $f(A)=[\max\{\vec{a}_1\},\dots,\max\{\vec{a}_n\}]$ and analysis the induced map between tangent spaces, but the $\max$ map here is kind of painful. Any thoughts?

Answer 1

It's not a constructive argument and I'm too shaky on my differential geometry to flesh it out completely off the top of my head, but I believe the following approach might be fruitful.

Fix a matrix $A$. For each $i$, let $m_i$ denote an index for which $|a_{i,m_i}| = \max(\vec a_i)$. Without loss of generality, suppose that $a_{i,m_i}>0$ for all $i$. Let $U$ denote the relatively open set $$ U = \{B : \text{for all }i, \max(\vec b_i) = b_{i,m_i}; b_{i,m_i} > |b_{ij}| \text{ for } j \neq m_i; \text{ and } b_{im_i} > a_{im_i}\}. $$ Consider the function $f:U \to \Bbb R^n$ given by $$ f(B) = [\max(\vec b_1), \dots, \max(\vec b_n)] = [b_{1,m_1},\dots,b_{1,m_n}]. $$ $f$ is the restriction of a linear and therefore differentiable, and (I think) the differential $df$ has full rank over $U$. It follows by [differential geometry argument of some kind] that for a sequence in $f(U)$ converging to the boundary point $f(A)$ of $f(U)$, there is a corresponding sequence in $U$ converging to $A$.