Solution: Consider,the set S of the nXn diagonal matrices with real entries with no zeros on the diagoal .Given A=[aij] in S,Since the determinant of a diagonal matrix is the product of its diagonal entries and aii =0 for all 1≤i≤n we have have that det(A)=a11a22....ann=0 which gives us that A∈GL(n,R).Hence S ⊂ GL(n,R).Given A=[aij] and B=[bij] in S we have that A and B are both diagonal matrices with real entries with no zeros on the diagonal which gives us thatAB=[aijbij].Since aii=0 and bii=0 for all 1≤i≤n,We have that aiibii=0 for all 1≤i≤n and therefore AB∈SSince In is a diagonal matrix with no zeros on its diagonal we have that In∈S.Given A=[aij] in S we have that B=A−1 is also a diagonal matrix with real entries such that bii=aii−1 for all 1≤i≤n.Thus since aii=0 for all 1≤i≤n,we have that bii=0 for all 1≤i≤n and therefore A−1∈S.Hence S is a subgroup of GL(n,R).Hence the set of the diagonal nXn matrices with real entries with no zeros on the diagoal is a subgroup of GL(n,R).Hence proved.
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