Question Number 143270 by SLVR last updated on 12/Jun/21
Answered by Olaf_Thorendsen last updated on 12/Jun/21
$$\mathrm{B}\left({i},{j}\right)\:\geqslant\:\mathrm{1} \\ $$$$\mathrm{Let}\:{c}_{{ij}} \:=\:\mathrm{B}^{\mathrm{2}} \left({i},{j}\right)\:=\:\underset{{p}=\mathrm{1}} {\overset{\lambda} {\sum}}\underset{{q}=\mathrm{1}} {\overset{\lambda} {\sum}}{b}_{{ip}} {b}_{{qj}} \\ $$$${c}_{{ij}} \:\geqslant\:\underset{{p}=\mathrm{1}} {\overset{\lambda} {\sum}}\underset{{q}=\mathrm{1}} {\overset{\lambda} {\sum}}\mathrm{1}×\mathrm{1}\:=\:\lambda \\ $$$$\mathrm{Let}\:{d}_{{ij}} \:=\:\mathrm{AB}^{\mathrm{2}} \left({i},{j}\right)\:=\:\underset{{p}=\mathrm{1}} {\overset{\lambda} {\sum}}\underset{{q}=\mathrm{1}} {\overset{\lambda} {\sum}}{a}_{{ip}} {c}_{{qj}} \\ $$$${d}_{{ij}} \:\geqslant\:\underset{{p}=\mathrm{1}} {\overset{\lambda} {\sum}}\underset{{q}=\mathrm{1}} {\overset{\lambda} {\sum}}\mathrm{1}×\lambda\:=\:\lambda^{\mathrm{2}} \\ $$$$\mathrm{tr}\left(\mathrm{AB}^{\mathrm{2}} \right)\:=\:\underset{{i}=\mathrm{0}} {\overset{\lambda} {\sum}}{d}_{{ii}} \:\geqslant\:\underset{{i}=\mathrm{0}} {\overset{\lambda} {\sum}}\lambda^{\mathrm{2}} \:=\:\lambda^{\mathrm{3}} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{answer}\:\left(\mathrm{a}\right)\::\:\lambda^{\mathrm{3}} \\ $$
Commented by SLVR last updated on 12/Jun/21
$${so}…{kind}\:{of}\:{you}\:{mr}.{Olaf}…{thank}\:{you} \\ $$$${so}\:{much} \\ $$