Wednesday, 15 April 2015

linear algebra - Sylvester rank inequality: operatornamerankA+operatornamerankBleqoperatornamerankAB+n


If A and B are two matrices of the same order n, then
rankA+rankBrankAB+n.





I don't know how to start proving this inequality. I would be very pleased if someone helps me. Thanks!



Edit I. Rank of A is the same of the equivalent matrix A=(Ir000). Analogously for B, ranks of A and B are r,sn. Hence, since rankAB=min, then r+s\leq \min\{r,s\} + n. (This is not correct since \operatorname{rank} AB \leq \min\{r,s\}.



Edit II. A discussion on the rank of a product of H_f(A) and H_c(B) would correct this, but I don't know how to formalize that \operatorname{rank}H_f(A) +\operatorname{rank}H_c(B) - n \leq \operatorname{rank}[H_f(A)H_c(B)].

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