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Question in regards to Permutation matrices, basis, and non linear functions

by SupaNerd
Tags: basis, functions, linear, matrices, permutation
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SupaNerd
#1
May6-05, 02:13 AM
P: 1
Hey thanks again, figured these questions out!
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kleinwolf
#2
May6-05, 04:06 AM
P: 309
Quote Quote by SupaNerd

1) Find a basis of R^4 that contains the vectors [ 1 2 3 4 ] and [ 0 1 0 1].
Note, they are both column vectors but I do not know how to orient them on this site.
let e_1 and e_2 be the 2 vectors you gave.

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.


2) Find a non linear function f : R^2 ---- > R^2
that still satisfies,
f(rv) = rf(v) for all r and all v.
if f is a function on R^2, then f is a function of 2 variables : f=f(x,y)...you gave only 1 variable for the condition on f.


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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