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{r,s}, then r+smin{r,s}+n. (This is not correct since rankABmin{r,s}.



Edit II. A discussion on the rank of a product of Hf(A) and Hc(B) would correct this, but I don't know how to formalize that rankHf(A)+rankHc(B)nrank[Hf(A)Hc(B)].

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