Question: Let a ~ i = x i + l i e i , where x i = j = 1 , j ! = i
Let a~i=xi+liei , where xi=j=1,j!=ikja~j lies in the subspace L spanned by a~1,...,a~ka~i,ei is in the orthogonal complement of L, ei=1,ei.a~i>0,li is a scalar. Prove det(A~TA~)=det(A~iTA~i)(liei.a~i) , where A~i=[a~1,...,a~i1,a~i+1,...,a~k] with a~iexcluded.
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