Multiplication of an orthogonal matrix and a skew symmetric matrix Let $A\in O(n;\mathbb{R})$ such that for every $1\leq i\leq n$, $1>a_{ii}>0$ and $a_{ii}\geq |{a_{ij}}|$ for $j\neq i$. Prove that there exists a skew symmetric matrix $B$ such that all diagonal elements of $AB$ are positive.
For $n=2$ the proof is simple: let $b_{12}=k$ and $b_{21}=-k$, then the diagonal elements of $AB$ are $-ka_{12}$ and $ka_{21}$. Since $a_{11},a_{22}>0$, $a_{21}$ and $a_{12}$ have different parities so a satisfying $k$ exists. I have also proved the case of $n=3$ by a similar parity argument, but I am not sure how to prove it in general. Thanks!
After some investigations I suspect that the parity argument couldn't be generalized, i.e., we really need the entries of $B$, not only the parities of the entries of $B$, for such $B$ to exists.
Edit: I believe some conditions here are redundant: over all we have $n(n-1)/2$ entries to choose for $B$ and we only need to satisfy $n$ inequalities, so as $n$ gets large it seems like it can always be done. But how to prove it though?

This is not always possible. E.g. when $n=5$, it is easy to generate by computer a symmetric orthogonal matrix $A$ such that $0<a_{ii}<1$ and $a_{ii}\ge|a_{ij}|$ using the following Octave/Matlab script:
n=5;
D=diag([ones(n-1,1); -1]);
for k=1:10000
  [U,S,V]=svd(2*rand(n,n)-1);
  A=U*D*U';
  if min(diag(A))>0 && max(diag(A))<1 && min(diag(A)'-max(abs(A)))>=0
    A, break;
  end
end

In this case, the trace of $AB$ is necessarily zero. Therefore, if $AB$ has a positive diagonal entry, it must have a negative diagonal entry too.
However, the desired $B$ exists if $A$ satisfies some additional conditions.
By permuting the rows and columns of $A$ if necessary, we may assume that the first $r$ diagonal entries of $A^2$ are equal to $1$ and the remaining $n-r$ diagonal entries are less than $1$. So, if $\mathbf a_j$ denotes the $j$-th column of $A$, the $j$-th row of $A$ must be equal to $\mathbf a_j^T$ when $j\le r$. Now the desired $B$ exists if one of the following sufficient conditions is satisfied:


*

*$r=0$. In this case, all diagonal entries of $A^2$ are smaller than $1$. Therefore, when $B=A^T-A$, all diagonal entries of $AB=I-A^2$ are positive.

*$r>0$ and the column space of the augmented matrix $M=\pmatrix{\mathbf a_{r+1}&\cdots&\mathbf a_n}$ contains some entrywise nonzero vector $\mathbf v=(v_1,v_2,\ldots,v_n)^T$ (e.g. when $M$ has an entrywise nonzero column or when $M$ has not any zero row). Let
$$
B=A^T(I+\epsilon \mathbf v\mathbf v^T)-(I+\epsilon \mathbf v\mathbf v^T)A.
$$
Then $AB=I-A^2+\epsilon(\mathbf v\mathbf v^T-A\mathbf v\mathbf v^TA)$. Since $\mathbf v\perp\mathbf a_j$ for all $j\le r$, the $j$-th diagonal entry of $AB$ is equal to $\epsilon v_j^2$ when $j\le r$, or $1-(A^2)_{jj}+\epsilon\times\text{some constant}$ when $j>r$. It follows that $AB$ has a positive diagonal when $\epsilon>0$ is small.

