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Let $A$ be a $n \times n$ real matrix of, say, rank $r$. Consider the matrix $$\max \{0,A\}$$ whereby each negative element of $A$ is set to $0$ and the non-negative elements are left unchanged. Is there anything known about by how much the rank of $A$ can increase by such a deformation?

I am typically thinking of the situation when $r \ll n$ and I am wondering if there are conditions known which if true then the new matrix also continues to have a rank far below $n$.

An analogy can be drawn to how under the Hadamard product rank is sub-multiplicative. Hadamard product of two low rank matrices can't be of rank arbitrarily high.

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  • $\begingroup$ Well, here's an easy example: if $A$ has rank one, then the deformed matrix has rank at most two. $\endgroup$
    – Gerry Myerson
    Commented Aug 1, 2017 at 23:41

2 Answers 2

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The deformation of a matrix of rank two can have full rank, e.g., $$\pmatrix{1&1&-1&-2&-3\cr1&2&0&-1&-2\cr1&3&1&0&-1\cr1&4&2&1&0\cr1&5&3&2&1\cr}$$

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  • $\begingroup$ Maybe this is an effect of the fact that here the rank i.e $2$ is already pretty close to the dimension i.e 5? $\endgroup$
    – gradstudent
    Commented Aug 2, 2017 at 0:02
  • $\begingroup$ I leave it to you, gradstudent, to show that 5 is a variable here, that is, the construction works for every $n$. $\endgroup$
    – Gerry Myerson
    Commented Aug 2, 2017 at 0:05
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    $\begingroup$ Okay. So for every n you can have a rank 2 matrix whose rank after max-0 will be n. $\endgroup$
    – gradstudent
    Commented Aug 2, 2017 at 0:08
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    $\begingroup$ Or maybe a bit easier to see: $$ \pmatrix{1 & 0 & -1 & -2 & -3\cr 2 & 1 & 0 & -1 & -2\cr 3 & 2 & 1 & 0 & -1\cr 4 & 3 & 2 & 1 & 0\cr 5 & 4 & 3 & 2 & 1\cr} = \pmatrix{1\cr 2\cr 3\cr 4\cr 5} \pmatrix{1 & 1 & 1 & 1 & 1} - \pmatrix{1\cr 1\cr 1\cr 1\cr 1\cr} \pmatrix{0 & 1 & 2 & 3 & 4}$$ $\endgroup$
    – Robert Israel
    Commented Aug 2, 2017 at 0:58
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    $\begingroup$ Lower triangular matrix with $1$'s on the diagonal. $\endgroup$
    – Robert Israel
    Commented Aug 2, 2017 at 2:37
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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

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

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.

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