I have a matrix $$ B=\begin{pmatrix} 0 & b_{12} & b_{13}\\ b_{21} & 0 & b_{23}\\ b_{31} & b_{32} & 0\\ \end{pmatrix} $$ with $b_{ij}<0$ $\forall i,j$. $B$ is not symmetric.
Could you suggest some sufficient conditions (possibly easily checkable) for $B$ to be positive definite or positive semidefinite?
On
For a symmetric matrix $B$, it follows from Sylverster's criterion that this is not possible. Positive definiteness requires that $b_{11} > 0$. The second leading principal minor is $-b_{21}b_{21} < 0,$ thus the matrix is also not positive semi-definite.
For non-symmetric matrices $B$, there is no general agreement on what a "positive definite non-symmetric matrix" is.
As SC Maree pointed out, the term "positive (semi-)definite" only applies to symmetric matrices.
But you might be interested in knowing whether $x^T B x \ge 0$ for all vectors $x$. Suppose this is true. Then in particular: $$ 0 \le (1,1,0)^T B\, (1,1,0) = b_{12}+b_{21}, $$ and $$ 0 \le (1,-1,0)^TB\,(1,-1,0) = - (b_{12}+b_{21}), $$ hence $b_{12} = - b_{21}.$ This contradicts your assumption that $b_{ij} < 0$ for all $i,j$.
Remark: Considering the vectors $(1,0,1), (1,0,-1)$ and $(0,1,1), (0,1,-1)$ also shows $b_{13} = - b_{31}$ and $b_{23} = - b_{32}.$ Hence for a three-dimensional matrix $B$ with zero diagonal to satisfy $x^T B x \ge 0$ for all $x$, it is necessary that $$ B = \begin{pmatrix} 0 & b_{12} & b_{13} \\ - b_{12} & 0 & b_{23} \\ - b_{13} & - b_{23} & 0 \end{pmatrix}. $$ You can check that in this case $x^T B x = 0$ for all $x$. So the condition is also sufficient.