Rank of block triangular matrix using linearly independent rows/columns

Question

Prove that $\operatorname{rank}\begin{pmatrix}A & X \\ 0 & B \\ \end{pmatrix}\ge \operatorname{rank}(A) + \operatorname{rank}(B)$ and the equality is attained if $X=0$ ($A, B, X \in M_{m, n} (\mathbb C) $) .
I know this is a pretty well-known relation, but I want to prove it using linearly independent rows or columns because all the other proofs I have seen seem way to advanced for me to understand.

Answer 1

This is a sketch. You should verify each claim using the definition of linear independence.

There exist $\text{rank}(A)$ columns of $A$ that are linearly independent so there exist $\text{rank}(A)$ columns of $\begin{pmatrix}A\\0\end{pmatrix}$ that are linearly independent.

There exist $\text{rank}(B)$ columns of $B$ that are linearly independent so there exist $\text{rank}(B)$ columns of $\begin{pmatrix}X\\B\end{pmatrix}$ that are linearly independent.

Together, you obtain $\text{rank}(A) + \text{rank}(B)$ columns of the full matrix that are linearly independent.

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Edit: response to comment

Your intuition is correct: somehow $X$ can help with getting larger linearly independent sets. I am not sure if I can give a precise description for why this is the case.

In any case here is an example. If $A = \begin{pmatrix}1 & 0 \\ 0&0 \end{pmatrix}$ and $B = \begin{pmatrix}0 & 0 \\ 1 & 1 \end{pmatrix}$ and $X = \begin{pmatrix}1&0\\0&1\end{pmatrix}$ then the inequality is strict.