Htykuut

I am new here!! I'm studying maths in Argentina.
I have a doubt abouts the next demostration:

Be $A∈M_{m,n}(R),Bin M_{p,l}(R),Cin M_{m,l}$ three blocks of matrix $D$, $Din M_{(m+p),(n+l)}(R)$, so
$D=begin{bmatrix}
A & C \
0 & B \
end{bmatrix}$, where $0$ is a block. Are there any formula that relate rank $D$ with rank $A,B,C$?

And later, for any block C, prove that: $rank(D) > rank (A)+rank (B)$

What I have tried is:

Exist $P,Q$ two matrix that $PDQ=begin{bmatrix}
I_{d} & 0 \
0 & 0 \
end{bmatrix}$ where $rank (D)=d$, doing the same for every block I have:
$A=begin{bmatrix} I_{a} & 0 \
0 & 0 \
end{bmatrix}$, $B=
begin{bmatrix}
I_{b} & 0 \
0 & 0 & \
end{bmatrix}$ and $C=begin{bmatrix}
I_{c} & 0 \
0 & 0 \
end{bmatrix}$, so $D=begin{bmatrix}
I_{a} & 0 & I_{c} & 0 \
0 & 0 & 0 & 0 \
0 & 0 & I_{b} & 0
end{bmatrix}$, Is my method correct? How I can prove it? And how I prove $rank D> rank A + rank C$

Thank you!

Your statement is not correct. The sign should be $ geqslant $, not $>$. Here's why.

The linearly independent columns of $A $ with a bunch of zeroes added at the bottom remain linearly independent. Hence, rank $D $ is at least rank $A$. The slightly tricky part is dealing with the column vectors whose upper components are comprised of columns of $ C $, lower, of $ B $; call these CB column vectors. In order to get the most linearly independent columns out of $B$, then as said before, $C$ should be zero; then rank $D$ = rank $A$ + rank$B$. However, $C $ may contain some vectors such that there are more linearly independent CB column vectors. For example, maybe all the $B$ vectors are $ begin{pmatrix} 1 \ 1 \ 1 end{pmatrix} $ thrice repeated, but the $C$ vectors are $begin{pmatrix} 1 \ 0 \ 0 end{pmatrix}, begin{pmatrix} 0 \ 1 \ 0 end{pmatrix}, begin{pmatrix} 0 \ 0 \ 1 end{pmatrix} $, in order. Whereas rank $B$ = 1, now rank$ D $ = rank $ A $ + 3 $ > $ rank $ A $ + rank $ B $.

Why not the reverse inequality? If you start with my $C$ column vectors (these are linearly independent), then no matter what you tack on to the bottom, there is no non trivial linear combination to make the $ C $ components of the CB column vectors zero by definition.