Neighborhood in orthogonal group

Question

Let $A\in O(n)$. Assume that $|a_{i,i}|\neq 1$ for every $i$. Prove that in every neighborhood of $A$ there exists $B\in O(n)$ such that $|b_{i,i}|>|a_{i,i}| \text{ for every } i \text{ and } |b_{i,j}|\leq |a_{i,j}| \text{ for every } i\neq j$.

I have thought about projecting $A+\epsilon I$ on $O(n)$ (wlog I assume $a_{ii}\geq 0$ here), but there doesn't seem to be a nice formula indicating whether this projection would work or not. Similarly I am not sure about Gram-Schmidt. Or maybe it's better to find a prove that doesn't use explicit constructions, but I don't know much about neighborhood in $O(n)$. Any suggestions?

Edit: the main difficulty I faced was that, if $a_{i,j}=0$ then we must force $b_{i,j}=0$, which prevents methods that involve 'small perturbation of every entry'.

Edit 2: As a counter-example is given below, I wonder whether the claim is true if $A$ is close to $I$, say if $A$ is closest to $I$ among all signed permutation matrices. (In the counter-example, $A\neq I$ itself is a signed permutation matrix).

Note that this claim is true is $A$ is close enough to $I$, as we can form a path $B_t=\text{exp}(t\log(A))$. Since the entries of $B_t$ is analytic on $t$, if $A$ is sufficiently close to $I$ every $B_t$ would satisfy the conditions for $t\in[0,1]$.

Answer 1

This is not always possible. As you have already discovered, a difficulty is that all off-diagonal zeroes in $A$ must be inherited by $B$. For instance, consider $$ A=\pmatrix{\color{red}{0}&\color{red}{1}&0\\ 0&\color{red}{0}&\color{red}{1}\\ \color{red}{1}&0&\color{red}{0}}. $$ Only the red elements are modifiable. As the rows and columns of $B$ are unit vectors, we must have $$ B=\pmatrix{c&s&0\\ 0&\pm c&\pm s\\ \pm s&0&\pm c} $$ for some real numbers $c$ and $s$ such that $c^2+s^2=1$. But then the first two columns of $B$ are orthogonal to each other only when $c=0$ or $s=0$ (so that $c=\pm1$). Therefore, $c$ cannot be a small nonzero number and $B$ does not exist in any small neighbourhood of $A$.

Answer 2

Consider the case when $A$ is in canonical form:

$$ A = \left[\begin{array}{cccccc}R_{1} & & & & & \\ & \ddots & & & & 0 \\ & & R_{K} & & & \\ & & & \pm 1 & & \\ & 0 & & & & \ddots \\ & & & & & \pm 1\end{array}\right] \qquad R_k = \left[\begin{array}{rr}\cos(\theta_k) & -\sin(\theta_k) \\ \sin(\theta_k) & \cos(\theta_k)\end{array}\right] $$

Here, the claim is obviously true: We can increase the size of the diagonal entries, while reducing the size of the off-diagonal entries by undoing the rotations. All that's left to is to extend this argument to the general case.