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Look here, my friend. How to prove the following equation? Or give a counter-example.

$$\text{rank}(A^+)\leq \text{rank}(A)$$ where $A$ has full rank, and $\text{rank}(A^+)$ represents the positive components of matrix $A$, e.g. $$\left[\begin{matrix}2 & -1.5 & 0\\-2.3 & 2 & 4.1\end{matrix}\right]^+=\left[\begin{matrix}2 & 0 & 0\\0 & 2 & 4.1\end{matrix}\right].$$ Thank you in advance.

guorui
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    If $A $ has full rank then evidently $\text{rank} B \leqslant \text{rank} A$ for any $B$ of the same size. – Minz Feb 28 '19 at 03:32
  • @Minz Exactly, I am now laughing and crying, and thank you. – guorui Feb 28 '19 at 03:43
  • @kingW3 No, there is one more condition "$A$ has full rank" than his question. – guorui Feb 28 '19 at 03:46
  • @Minz By the way, if you add this comment to formally answer this question below, I'd like to accept this great answer. – guorui Feb 28 '19 at 06:52

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Counter example: $A = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$.

Anthony Ter
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  • There is one more constraint "A has full rank". As @Minz commented above, the proof is obvious. ^_^ – guorui Feb 28 '19 at 06:59