# Rank of block triangular matrix using linearly independent rows/columns

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.

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.

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.

• Thank you! What I do not understand is why we get that the rank of the full matrix is $\ge$ and not equal to $rank A+rank B$. Commented Jan 22, 2019 at 18:11
• Is it because X may have more linearly independent columns? This would explain why the equality is attained when $X=0$. Commented Jan 22, 2019 at 18:28
• @JustAnAmateur See my edit Commented Jan 22, 2019 at 18:42
• Thank you very much, I had been struggling with proving this result for some time. Commented Jan 22, 2019 at 18:47