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I wanted my proof verified. $\dim(A)=m,n$ and same for $B$.

We know that size of $A+B$ is $m\times n$. This means that maximum rank for $A+B$ is $m$.

Consider that $r_1, r_2,r_3...r_m$ are rows of $A$ and $r'_1, r'_2..., r'_m$ are rows of $B$. If $\operatorname{rank}(A)=k$ and $\operatorname{rank}(B)=l$ such that $k+l>m$ then the inequality holds trivially. In the other case$(k+l\le m)$ let's form the set $\{r_1,r_2...r_k, r'_1, r'_2,..., r'_l \}$ this set contains $k+l$ elements and they can be used to create all the rows of $A+B$. If this set is independent then we have that equality holds. But if this set is linearly dependent then we can say that it will contain less than $k+l$ elements and since it could still be used to recreate all rows of $A+B$ we can say that rank of $A+B$ is less that $k+l$. Hence the inequality holds.

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One problem with the proof, where you say

We know that size of $A+B$ is $m\times n$. This means that maximum rank for $A+B$ is $m$.

The maximum rank for $A + B$ may not be $m$, in particular if $n < m$, in which case, $n$ is an upper bound for matrices of size $m \times n$. You should also be more clear with the term "maximum rank", as to what exactly you're maximising over. In this case, I think the phrase "$m$ is the maximum rank of matrices with dimensions $m \times n$" is more precise.

However, I believe you're just using the fact that $\operatorname{rank}(A + B) \le m$, which is indeed true.

You seem to be missing one of the crucial cases for your argument, in particular, $k + l < m$. I think your existing arguments can be adapted. However, I think the primary issue is that you don't really need to refer to $m$ at all; it's just holding you back.

Just take $\lbrace r_1, \ldots, r_k \rbrace$ and $\lbrace r'_1, \ldots, r'_l\rbrace$ as before, and consider their union. Simply note that the span of this union spans the rowspace of $A + B$ (and possibly more!). The dimension of this span is therefore larger than or equal to $\operatorname{rank}(A + B)$. However, the dimension is smaller than or equal to $\operatorname{rank}A + \operatorname{rank}B$. Therefore, $$\operatorname{rank}(A + B) \le \operatorname{rank}A + \operatorname{rank}B,$$ as required.

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  • $\begingroup$ Thanks this is much better. I didn't need to use m. However I did try to include that case, in the question the other case i refer to is $k+l\le m$. I didn't state it explicitly. But your argument is much better than what I came up with. $\endgroup$ – Sonal_sqrt Jun 13 '18 at 3:49

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