Rank of the sum of two matrices

Question

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

Answer 1

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 $$\operatorname{im} M\cap \operatorname{im} N=\{0\}$$