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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.

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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 $\operatorname{rank}A = \operatorname{rank}A_{11}+\operatorname{rank}A_{22}+\cdots +\operatorname{rank}A_{kk}$.

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    $\begingroup$ The induction idea is good, but you have not proven the $k = 2$ case. $\endgroup$ Commented Feb 3, 2019 at 6:19

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