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Let $M = \pmatrix{A & B\\ C& D}$, where $A$ is an all-one matrix. From Section 3 of Nisan & Wigderson$\color{magenta}{^\star}$,

$$\operatorname{rk} (B) + \operatorname{rk} (C) \le \operatorname{rk} (M) + 1$$

Why is this true?

$\color{magenta}{\star}$ Noam Nisan, Avi Wigderson, On rank vs. communication complexity.

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If $A$ was all-0 matrix, we would have $rk(M)\geqslant rk(B)+rk(C)$ (take $rk(C)$ linearly independent columns of $C$ and $rk(B)$ linearly independent columns of $B$, the corresponding $rk(B)+rk(C)$ columns of $M$ are linearly independent), and replacing $A$ to all-1 changes the rank by at most 1.

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  • $\begingroup$ why replacing $A$ to all-one changes the rank by at most 1? $\endgroup$ Commented Jul 22 at 15:30
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    $\begingroup$ @Connor Think about the column span. If we make $A$ all $1$'s but also add a new column that is a bunch of $1$'s (equal to dimension of $A$) followed by a bunch of $0$'s (equal to dimension of $C$), then the column span of this new matrix certainly contains the column span of the matrix when $A$ was all $0$'s. $\endgroup$ Commented Jul 22 at 15:58
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    $\begingroup$ Because rank of the sum does not exceed the sum of ranks, thus, rank-1 peturbation (and replacing $A$ to all-0 is adding a matrix of rank 1) does not change the rank by more than 1 $\endgroup$ Commented Jul 22 at 20:07

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