TITLE: Rank of the sum of two matrices
QUESTION [4 upvotes]: I have two square matrices $M$ and $N$ such that $M^2=M$ , $N^2=N$ and $MN=NM=0$. I'd like to prove that $\operatorname{rank} (M+N)=\operatorname{rank} (M)+\operatorname{rank} (N)$. I know that $\operatorname{rank} (M+N)\leq \operatorname{rank} (M)+\operatorname{rank} (N)$. But how will i prove $\operatorname{rank} (M+N)\geq \operatorname{rank} (M)+\operatorname{rank} (N)$? Thanks for any help

REPLY [6 votes]: Since $M^2=M$ and $N^2=N$ hence $M$ and $N$ are two projectors hence
$$\mathbb R^n=\ker M\oplus \operatorname{im} M=\ker N\oplus \operatorname{im}N$$
moreover $MN=0=NM$ so
$$\operatorname{im}N\subset \ker M  \quad\text{and}\quad \operatorname{im} M\subset \ker N $$
now let $y\in \operatorname{im} M$ so $y=Mx$ but $x=x_1+x_2$ where $x_1\in \ker N$ and $x_2\in \operatorname{im} N\subset\ker M$ so 
$$y=Mx=Mx_1=(M+N)x_1$$
hence 
$$\operatorname{im} M\subset \operatorname{im} (M+N)$$
we prove by similar method that
$$\operatorname{im} N\subset \operatorname{im} (M+N)$$
hence
$$\operatorname{im} M+\operatorname{im} N\subset \operatorname{im} (M+N)$$
finaly the other inclusion is simple so we have the equality and the result follows since we have
$$\operatorname{im} M\cap \operatorname{im} N=\{0\}$$
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