Rank of a block lower triangular matrix 
Let $A$ be a $k \times k$ block lower triangular matrix
$$A = \left[\begin{matrix}
A_{11} & 0 & \cdots & 0\\
A_{21} & A_{22} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
A_{k1} & A_{k2} & \cdots & A_{kk}
\end{matrix}\right]$$
Prove that
$$\mbox{rank} A \geq \mbox{rank} A_{11} + \mbox{rank} A_{22} + \cdots + \mbox{rank} A_{kk}$$
and that equality holds if $A_{ij} = 0, i > j$.

I have tried to solve this by induction but I am not getting the proof for this.
 A: Ans: We prove this by induction on k.
For k = 1, $ A = \left[\begin{matrix}
A_{11}
\end{matrix}\right] $. Then $\operatorname{rank}A = \operatorname{rank}A_{11}$, result holds.
For k = 2, $A = \left[\begin{matrix}
A_{11}& O\\
A_{21}& A_{22}
\end{matrix}\right]$. Now, to prove $\operatorname{rank}A \eqslantgtr \operatorname{rank}A_{11}+\operatorname{rank}A_{22}$.
Suppose $\operatorname{rank}A < \operatorname{rank}A_{11}+\operatorname{rank}A_{22}$.
Putting $A_{21} = O$, we get $A = \left[\begin{matrix}
A_{11}& O\\
O & A_{22}
\end{matrix}\right]$. 
Then $\operatorname{rank}A = \operatorname{rank}A_{11}+ \operatorname{rank}A_{22}$, which is a contradiction. Thus $\operatorname{rank}A \eqslantgtr \operatorname{rank}A_{11}+\operatorname{rank}A_{22}$.
Asuuming the result for k-1, we prove the result for k.
We partition the matrix A as
$\left[\begin{matrix}
B_{11}& O\\
B_{21}& A_{kk}
\end{matrix}\right]$ where $B_{11} = \left[\begin{matrix}\\
A_{11} & 0 & \cdots & 0\\
A_{21} & A_{22} & \cdots & 0\\
\vdots & \vdots & \ddots & \\
A_{(k-1) 1} & A_{(k-1)2} & \cdots & A_{(k-1)(k-1)}. 
\end{matrix}\right]$ and $B_{21} = \left[\begin{matrix}
A_{k1}&A_{k2}& \cdots &A_{k(k-1)}
\end{matrix}\right]$
Then $\operatorname{rank}A \eqslantgtr \operatorname{rank}B_{11}+\operatorname{rank}A_{kk}$ (By k = 2 case)
By induction hypothesis, $\operatorname{rank}B_{11}\geq \operatorname{rank}A_{11}+ \operatorname{rank}A_{22}+\cdots +\operatorname{rank}A_{(k-1)(k-1)}$
$\therefore \operatorname{rank}A \geq \operatorname{rank}A_{11}+ \operatorname{rank}A_{22}+\cdots +\operatorname{rank}A_{(k-1)(k-1)}+\operatorname{rank}A_{kk}$.
If $A_{ij} = 0, i > j$ then $A = \left[\begin{matrix}
A_{11}&0&\cdots&0\\
0&A_{22}&\cdots&0\\
\vdots&\vdots&\ddots\\
0&0&\cdots&A_{kk}
\end{matrix}\right]$.
Then, clearly $\operatorname{rank}A = \operatorname{rank}A_{11}+\operatorname{rank}A_{22}+\cdots +\operatorname{rank}A_{kk}$. 
