In some cases, the rank is preserved under thresholding. For example, let

$$\rm A := \begin{bmatrix} 1\\ 0\\-1\end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}^\top = \begin{bmatrix} 1 & 1 & 1\\ 0 & 0 & 0\\-1 & -1 & -1\end{bmatrix}$$

Note that $\rm A$ is a rank-$1$ matrix. Thresholding $\rm A$, we obtain another rank-$1$ matrix

$$\max \{ \mathrm O_3, \mathrm A \} = \begin{bmatrix} 1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix} = \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}^\top$$

Thinking of the thresholding operation in terms of the [Hadamard product][1]

$$\max \{ \mathrm O_3, \mathrm A \} = \underbrace{\begin{bmatrix} 1 & 1 & 1\\ 0 & 0 & 0\\-1 & -1 & -1\end{bmatrix}}_{= \mathrm A} \circ \underbrace{\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 0 & 0 & 0\end{bmatrix}}_{=: \mathrm B}$$

where $\rm B$ is a binary matrix that contains information pertaining to the signs of the entries of $\rm A$. In this case, $\rm B$ is also a rank-$1$ matrix. Since

$$\mbox{rank} (\mathrm A \circ \mathrm B) \leq \mbox{rank} (\mathrm A) \cdot \mbox{rank} (\mathrm B)$$

and $\mbox{rank} (\mathrm A) = \mbox{rank} (\mathrm B) = 1$, the rank does not increase under thresholding. Since $\mathrm A \circ \mathrm B \neq \mathrm O_3$, we conclude that the rank is actually preserved.


  [1]: https://en.wikipedia.org/wiki/Hadamard_product_(matrices)