
#1
Mar1512, 08:55 AM

P: 23

1. The problem statement, all variables and given/known data
If set A={u,v,w} ⊂ R^n is linearly independent, is B={uv, u+w, v+w}⊂ R^n linearly independent? 2. Relevant equations 3. The attempt at a solution Since A is linearly independent, there exist no all nonzero scalars a1, a2, a3 such that a1*u+a2*v+a3*w=0. Or I can say that since A is linearly independent, a1=a2=a3=0. Then to determine whether B is linearly indepedent, I think I need to determine whether b1=b2=b3=0 for b1*(uv)+b2*(u+w)+b3*(v+w)=0. But if I do so, I find no connection between a1, a2, a3 and b1, b2 ,b3. How can I complete this proof? 



#2
Mar1512, 09:11 AM

Mentor
P: 21,018





#3
Mar1512, 09:31 AM

P: 23

if u=(u1, u2, ....un), v=(v1, v2, ...vn), w=(w1,w2...wn) then b1(uv)+b2(u+w)+b3(v+w)=0 becomes b1(u1v1)+b2(u1+w1)+b3(v1+w1)=0 b1(u2v2)+b2(u2+w2)+b3(v2+w2)=0 . . . b1(unvn)+b2(un+wn)+b3(vn+wn)=0 Then how should i find b1, b2 and b3? Also I do not quite get what you means by 'Your will need to use the fact that a1*u + a2*v + a3*w = 0 has only the nontrivial solution for the three constants' 



#4
Mar1512, 12:55 PM

Mentor
P: 21,018

determining linear independenceFrom that equation you should be able to argue that u  v, u + w, and v + w are also linearly independent. 



#5
Mar1612, 01:30 PM

P: 7

Maybe I did something wrong, correct me please if I did, but I think B is linearly dependent. A linear dependence relationship would be b1=1, b2=1, b3=1, or any multiple of it. I uploaded a pdf file showing how I found this. The way this problem was worded led me to think B should be linearly independent, or I would've done it quicker.




#6
Mar1612, 02:05 PM

Mentor
P: 21,018





#7
Mar1612, 02:11 PM

HW Helper
P: 1,932

I don't think it's showing off but I took one look at that and thought hey second element minus first equals third, there is linear dependence without any if coming into it. IOW I think your conclusion is right.




#8
Mar1612, 04:12 PM

P: 37

I personally think the easiest way to solve these problems is calculating the determinant of a matrix whose columns are the vectors. I shall explain this better: A given vector (u,v,w) is an element of a tridimensional vector space (for example ℝ^3). A "basis" of this vector space is C= {(u,0,0);(0,v,0);(0,0,w)} = { e1, e2, e3 } Now we are given a set of vectors B= {uw,u+v,v+w} which can also be expressed this way: B= { (e1e3), (e1+e2), (e2+e3) } or, as a combination of vectors of C: B={ (1,0,1), (1,1,0), (0,1,1)} Now take the last three vectors and form a matrix with them, columns or rows doesn´t matter (if the determinant of a given matrix is ≠ 0, then the determinant of its transpose also is ≠ 0). If M is your matrix, calculate det(M) and you will get det(M)=0, which means that the vectors forming that matrix are linearly dependant. If det(M)≠0, then they would be linearly independant. Related to this is the concept of "rank". The rank of a given matrix indicates how many vectors (rows or columns) are linearly independant. If your matrix is size "n x n" and its determinant is unequal to cero, then its rank would be "n". In your problem rank(M)=2, meaning that there are only 2 independant vectors in set B. Knowing the rank of a set of vectors is very useful because it tells you whether you have "repeated information" or not. For example, if you have a system of 10 equations, assimilate each equation to a vector, and after calculating it turns out that the rank of the matrix formed by them is only 4, that means that you can throw away 6 equations and end up with an equivalent system (meaning that the solutions of the first and the ones of the second system are exactly the same). Of course you have to carefully select the equations you throw away, making sure that they are a linear combination of the equations you already have. Greetings! 



#9
Mar1712, 07:58 PM

P: 13

I see!



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