prove that those vectors form a basis

prove that those vectors form a basis

Postby Guest » Wed Sep 30, 2020 8:26 am

4. Given v1 = (1, 0, 1, 2)T
, v2 = (0, 1, 1, 0)T
, v3 = (−1, 2, 1, 0)T
, v4= (0, 0, 1, 0)T
.
(i) Prove that v1, v2, v3, v4 form a basis of R4
;

I have no idea because i thought basis are pivot columns like (1 0 0 0),(0 1 0 0), but the above are not pivot columns.
Thanks a lot
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Re: prove that those vectors form a basis

Postby Guest » Tue Oct 20, 2020 2:53 pm

"Pivot Columns" are a concept in matrices but there are no matrices in this so I don't know what you could mean.
Are you being asked to show that a given set of vectors is a "basis" for a vector space without having learned what a "basis" is?

A "basis" for a vector space (here [tex]R^4[/tex]) is a set of vectors that
1) span the space (every vector can be written as a linear combination of those vectors).
2) the vectors are independent (the only linear combination equal to the 0 vector has all coefficients 0).
Together those say that every vector can be written as a linear combination of those vectors in one and only one way.

So to show that (0, 1, 1, 2), (0, 1, 1, 0), (-1, 2, 1, 0), and (0, 0, 1, 0) (I dropped the "T" because it really doesn't matter if the vectors are written as rows or columns) we must show
1) The vector (p, q, r, s)= a(0, 1, 1, 2)+ b(0, 1, 1, 0)+ c(-1, 2, 1, 0)+ d(0, 0, 1, 0)= (-c, a+ b+ c, a+ b+ c+ d, 2a) for some numbers a, b, c, and d. So we need to show that, for any p, q, r, and s, we can solve -c= p, a+b+c= q, a+ b+ c+ d= r, and 2a= s for a, b, c, and d. Obviously, from the first equation, c= -p and, from the last, a= s/2. Then a+ b+ c= s/2+ b- p= q so b= p+ q- s/2. a+ b+ c+ d= s/2+ p+ q- p+ d= q+ s/2+ d= r so d= r- q- s/2.
That is, any vector (p, q, r, s)= (s/2)(0, 1, 1, 2)+ (p+ q- s/2)(0, 1, 1, 0)- p(-1, 2, 1, 0)+ (r- q- s/2)(0, 0, 1, 0).
So this set of vectors spans the space.

2) To show that the set of vectors is independent, we show that if a(0, 1, 1, 2)+ b(0, 1, 1, 0)+ c(-1, 2, 1, 0)+ d(0, 0, 1, 0)= (0, 0, 0, 0) the we must have a= b= c= d= 0.
To do that write a(0, 1, 1, 2)+ b(0, 1, 1, 0)+ c(-1, 2, 1, 0)+ d(0, 0, 1, 0)= (0, a, a, 2a)+ (0, b, b, 0)+ (-c, 2c, c, 0)+ (0, 0, d, 0)= (-c, a+ b+ 2c, a+ b+ c+ d, 2a)= (0, 0, 0, 0).
We must have -c= 0, a+ b+ 2c= 0, a+ b+ c+ d= 0, and 2a= 0. Now the first and last equations givw c= 0 and a= 0. Then a+ b+ 2c= b= 0 a+ b+ c+ d= b+ d= d= 0. We must have a= b= c= d= 0 so this set is independent.

Therefore this is a basis for [tex]R^4[/tex].

(It can also be shown that a basis for a vector space of finite dimension, n, must contain n vector so a basis for [tex]R^n[/tex] has three properties:
1) It spans the space.
2) The vectors are independent.
3) It contains n vectors.

Further any two of those is sufficient to prove the third!
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