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

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A basis for R^4 can be formed by including the vectors [1, 2, 3, 4] and [0, 1, 0, 1], along with additional orthogonal vectors like [1, 0, 0, 0] and [0, 0, 1, 0], ensuring linear independence. For non-linear functions f: R^2 → R^2 that satisfy f(rv) = rf(v), the function must be defined in terms of two variables, not just one. The application of permutation matrices on a vector results in a rearrangement of its elements, and the composition of two permutation matrices yields another permutation matrix. This discussion highlights the importance of understanding linear independence, basis formation, and the properties of permutation matrices in linear algebra.
SupaNerd
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Hey thanks again, figured these questions out!
 
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SupaNerd said:
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)
 


Great to hear that you were able to figure out your questions on permutation matrices, basis, and non-linear functions! Permutation matrices are square matrices that represent the rearrangement of rows and columns in a matrix. They are useful in linear algebra for solving equations and finding eigenvalues.

A basis is a set of linearly independent vectors that can be used to express any vector in a vector space. It serves as a starting point for understanding the structure of a vector space and can help with solving problems in linear algebra.

Non-linear functions are functions that do not follow a straight line and cannot be expressed as a linear combination of variables. They are commonly used in fields such as economics, physics, and computer science to model complex relationships between variables.

I hope this helps clarify these concepts for you. Keep up the good work in your studies!
 
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