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Let $M$ be a real matrix of rank $r$ (and let us set $M=UV^T$, with $U,V^T\in\mathbb{R}^{n\times r}$, to fix the notation).

Let $|M|$ be the matrix obtained by taking the absolute value of each entry of $M$. Clearly $\operatorname{rk} |M|$ can be much smaller than $r$ --- take for instance Hadamard matrices.

However, what about the other direction? Is there a way to bound $\operatorname{rk} |M|$ from above in terms of $r$?

If I take random $U$ and $V$ with $n=200$, $r=2$, numerically $\operatorname{rk} |M|$ seems to be between 120 and 150 --- so definitely not as low as $r$ but also suspiciously far from being full-rank.

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up vote 6 down vote accepted

I do not think much/anything can be done.

Let us leave the simple special cases of rank $M$ equal $0$ or $1$ aside.

So, an example of a $n$ times $n$ rank two matrix $M$ such that the rank of $|M|$ is full:

Take the two vectors $e=(1, \dots, 1)$ and $u = (0, -1, -2, \dots, -(n-1))$. Consider the matrix $M$ formed by $e$ and $je + u$ for $j=0, \dots n-2$.

The absolute value $|M|$ has full rank since the line for any $j\ge 1$ is $je+u + v_j$ where $v_j = (2 \max(0, (i-1)-j) )_i$.` and thus has exactly the first $j+1$ coordinates equal to $0$. So, we get $e$, and $u= - |u|$ and all the $v_j$ for $j=1, \dots, n-2$ in the spanned space, and these $n$ are independent.

Variations of this should give (all?) kinds of intermideate phenomena.

(Edit: slight change and explanation; perhaps the orginal would also work but the present example seems clearer and was the 'real' original, which I thought I should modify while typing for some dubious reasons. Sorry for the edit-noise.)

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Another example of $M$ with rank $2$ and $|M|$ with full rank is given by $M_{i,j}=i-j$. – Mikael de la Salle Feb 28 '12 at 15:34
And an example of $M$ of size $n$ with rank $2$ with $|M|$ of arbitrary rank $2 \leq r \leq n$ is obtained by taking a surjective map $\sigma:\{1,\dots,n\} \to \{1,\dots,r\}$ and setting $M_{i,j} = \sigma(i) - \sigma(j)$. – Mikael de la Salle Feb 28 '12 at 16:50

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