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I found a mathematical note by George Marsaglia entitiled "Bounds for the rank of the sum of two matrices", where he proves the following result.

Let $A_1$ and $A_2$ be two complex matrices of the same size. Then $$ \operatorname{rk}(A_1 + A_2) \geq \operatorname{rk}(A_1) + \operatorname{rk}(A_2) - \dim(\mathcal{R}_1 \cap \mathcal{R}_2) - \dim(\mathcal{C}_1 \cap \mathcal{C}_2),$$

where $\operatorname{rk}()$ denotes the rank, $\mathcal{R}_i$ (resp. $\mathcal{C}_i$) is the rowspace (resp. columnspace) of $A_i$, for $i = 1, 2$.

How does one extend this formula to a sum of, say, $n$ matrices $A_1, \dots, A_n$?

I conjecture that: $$ \operatorname{rk}(A_1 + \dots + A_n) + \sum_{i=1}^n \operatorname{rk}(A_i) \geq \operatorname{rk}(A_1, \dots, A_n) + \operatorname{rk}\begin{pmatrix} A_1 \\ \vdots \\ A_n \end{pmatrix}.$$

The first, resp. second, term on the RHS is the rank of the matrix having the $A_i$ stacked next to each other horizontally, resp. vertically.

I have run some numerical simulations by generating, each time, a large number (usually $1000$) of collections of $n$ (pseudo-)random matrices with integer entries and specified ranks, for some low values of $a$, $b$, $n$ and some choices of $n$ ranks. I could not yet find any counterexample this way. I am starting to think this could be true.

Note: I had written another conjecture initially, but the original conjecture was false. A counterexample can be obtained by taking $A_1 = (1, 0)$, $A_2 = (0, 1)$ and $A_3 = (-1, -1)$. In order not to clutter this post, I deleted the false conjecture and kept the one above, which may still possibly be true (or not).

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Yes. Let $A_i=U_i|A_i|$ denote the polar decomposition and put $X=(U_1|A_1|^{1/2},\ldots,U_n|A_n|^{1/2})$ and $Y=(|A_1|^{1/2},\ldots,|A_n|^{1/2})^T$. Thus $\sum A_i = XY$ and \begin{align*} \mathop{\mathrm{rk}} XY &= \mathop{\mathrm{rk}}Y-\dim(\ker X \cap\mathop{\mathrm{ran}}Y)\\ &= \mathop{\mathrm{rk}}X+\mathop{\mathrm{rk}}Y - \dim((\ker X)^\perp+\ker X\cap\mathop{\mathrm{ran}}Y). \end{align*} Observe that $\mathop{\mathrm{rk}}X=\mathop{\mathrm{rk}} {(A_1,\ldots,A_n)}$, $\mathop{\mathrm{rk}} Y=\mathop{\mathrm{rk}}{(A_1,\ldots,A_n)^T}$, and that $$(\ker X)^\perp + \mathop{\mathrm{ran}}Y\subset\bigoplus_i (\ker|A_i|)^\perp = \bigoplus_i \mathop{\mathrm{ran}}|A_i|$$ has dimension $\le\sum_i\mathop{\mathrm{rk}} A_i$.

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  • $\begingroup$ Your answer assumes that the matrices are square, but OP is interested also in the rectangular case, as can be seen from the counterexample at the end. $\endgroup$ Commented Jul 17, 2024 at 9:20
  • $\begingroup$ @Federico Poloni: I do not assume that matrices are square. $\endgroup$ Commented Jul 17, 2024 at 9:32
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    $\begingroup$ I realized that I used polar decomposition because of my operator theoretic background. It was equally OK to use the more simpler decomposition $A_i=A_iP_i$, where $P_i$ is the orthogonal projection onto the row space of $A_i$. $\endgroup$ Commented Jul 17, 2024 at 10:00
  • $\begingroup$ It took me a while to understand your second equality for $\operatorname{rk}(XY)$! You skipped a step, in the exposition, which was probably obvious to you, but anyway. Here it is. $\operatorname{rk}(X) = \dim((\operatorname{ker} X)^\perp)$. We also have that $(\operatorname{ker} X)^\perp$ and $\operatorname{ker} X \cap \operatorname{ran} Y$ have trivial intersection. $\endgroup$ Commented Jul 17, 2024 at 19:23
  • $\begingroup$ Can you please provide a little more details? For example, why is $\operatorname{rk}(X)$ equal to the rank of the horizontally concatenated $A_i$, and a similar question for $\operatorname{rk}(Y)$? If you could perhaps edit it for rectangular matrices, like you wrote in a comment, then that would be great. I am just worried that we are not multiplying the $A_i$ by the same matrix (either from the left or from the right). So wouldn't this possibly affect the rank? $\endgroup$ Commented Jul 18, 2024 at 0:06

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