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Question Number 143270 by SLVR last updated on 12/Jun/21

Answered by Olaf_Thorendsen last updated on 12/Jun/21

$$\mathrm{B}(i,j)⩾1$$$$\mathrm{Let}{c}_{ij}={\mathrm{B}}^2(i,j)=\underset{p=1}{\sum ^{\lambda }}\underset{q=1}{\sum ^{\lambda }}{b}_{ip}{b}_{qj}$$$$c_{ij}⩾\underset{p=1}{\sum ^{\lambda }}\underset{q=1}{\sum ^{\lambda }}1\times 1=\lambda$$$$\mathrm{Let}{d}_{ij}={\mathrm{AB}}^2(i,j)=\underset{p=1}{\sum ^{\lambda }}\underset{q=1}{\sum ^{\lambda }}{a}_{ip}{c}_{qj}$$$$d_{ij}⩾\underset{p=1}{\sum ^{\lambda }}\underset{q=1}{\sum ^{\lambda }}1\times \lambda ={\lambda }^2$$$$\mathrm{tr}\left({\mathrm{AB}}^2\right)=\underset{i=0}{\sum ^{\lambda }}{d}_{ii}⩾\underset{i=0}{\sum ^{\lambda }}{\lambda }^2={\lambda }^3$$$$$$$$\Rightarrow \mathrm{answer}\left(\mathrm{a}\right):{\lambda }^3$$

Commented by SLVR last updated on 12/Jun/21

$$so...kindofyoumr.Olaf...thankyou$$$$somuch$$

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