2
$\begingroup$

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.

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

2 Answers 2

Reset to default
2
$\begingroup$

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

Improve this answer
$\endgroup$
7
  • $\begingroup$ 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
    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
    Aug 2, 2017 at 0:05
  • 1
    $\begingroup$ Okay. So for every n you can have a rank 2 matrix whose rank after max-0 will be n. $\endgroup$
    – gradstudent
    Aug 2, 2017 at 0:08
  • 3
    $\begingroup$ 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
    Aug 2, 2017 at 0:58
  • 1
    $\begingroup$ Lower triangular matrix with $1$'s on the diagonal. $\endgroup$
    – Robert Israel
    Aug 2, 2017 at 2:37
1
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.