Prove if $AB = 0 = BA$ implies that $\text{rank}(A+B) = \text{rank }A + \text{rank }B$
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3What about $A=B=\begin{bmatrix}0&1\0&0\end{bmatrix}$? Then $\mathrm{rank}(A+B)=\mathrm{rank}(A)=\mathrm{rank}(B)=1$. – stewbasic Oct 06 '16 at 05:42
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That is only one particular case. – Hector1993 Oct 06 '16 at 05:48
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Yes but that particular case is a counterexample AFAICT, so your statement is not true for all $A$ and $B$. Do you have some additional conditions on $A$ and $B$? – stewbasic Oct 06 '16 at 23:10
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$A$ and $B$ are Hermitian. – Hector1993 Oct 07 '16 at 03:56
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This implies $A$ and $B$ are both diagonalizable, so you can use http://math.stackexchange.com/a/618319/197161 – stewbasic Oct 07 '16 at 06:18