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Question in regards to Permutation matrices, basis, and non linear functions 
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#1
May605, 02:13 AM

P: 1

Hey thanks again, figured these questions out!



#2
May605, 04:06 AM

P: 309

obviously e_1 and e_2 are lin. indep, so it's possible...just take orthogonal vectors to the previous ones. But you could just use : e_3=[1 0 0 0] and e_4=[0 0 1 0]...it's obvious that e_1,e_2,e_3 are pairwise orthogonal, hence lin. indep.....combining e_2,e_3,e_4 as you want will give you always a vector of the type [c b d b] hence you cannot express e_1 as a combintation of the others, so the e_i are lin. indep....clearly it's a generating subset of R^4...so it's a basis. 3) I think it's clear that applying a permutation matrix on an n vector a=[a1...an] corresponds to a permutation of the elements of the vector ? b=M1*a Apply a second permutation matrice on b : c=M2*b since c is permutation of b and b a permutation of a, then since the permuation are building a group, c is a permutation of a..hence there exists a permutation matrix M3 such that c=M3*a...but from above c=M2*M1*a....for all a...then M3=M2*M1... (This is not a good proof but I cannot think of a brute force computation of direct multipl of two perm. matrices defined as having a 1, and only 1, in every row and column) 


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