# rank of block triangular matrix and its relation to the rank of its diagonal blocks

Prove that $$rank\begin{pmatrix}A & X \\ 0 & B \\ \end{pmatrix}\ge rank(A) + rank(B)$$ where $$A,B\in \mathbb C^{m \times m}$$.

I know the intuition behind it (i.e. maximal independent rows, etc.), but I am looking for a formal proof. I have tried QR decomposition of A and B, then broke the block triangular matrix up into a matrix product and used the inequality rank(AB)<= min{rank(A), rank(B)}, but have not made any great progress. One fact I thought may help was, if A=QR, where Q is unitary, then rank(A)=rank(R). Again I thought this would help but it did not. Any help on this problem would be much appreciated.

## 1 Answer

Let $a = \text{rank}(A)$ and $b=\text{rank}(B)$. Consider a set of $a+b$ columns $C_1, \ldots, C_{a+b}$ of your matrix, where $C_1, \ldots, C_a$ correspond to linearly independent columns of $A$ and $C_{a+1}, \ldots, C_{a+b}$ correspond to linearly independent columns of $B$. Suppose $\sum_{i=1}^{a+b} s_i C_i = 0$ where $s_1, \ldots, s_{a+b}$ are scalars. Because those $b$ columns of $B$ are linearly independent, we must have $s_{a+1} = \ldots = s_{a+b} = 0$. And then because those $a$ columns of $A$ are linearly independent, $s_1 = \ldots = s_a = 0$.