Prove that v2 is the only element in W_1\cap W_2.

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Subspace
transgalactic
Messages
1,386
Reaction score
0
V is a vector space on field F and there is a seriesv=(v_1,v_2,v_3,v_4)
which is independent on V
W_1=sp(v1,v2)
W_2=sp(v2,v3)
of V
prove that
<br /> W_1\cap W_2=sp(v2) <br />
its obvious v2 exists in both groups .
how am i supposed to prove it
??
 
Physics news on Phys.org
transgalactic said:
V is a vector space on field F and there is a seriesv=(v_1,v_2,v_3,v_4)
which is independent on V
W_1=sp(v1,v2)
W_2=sp(v2,v3)
of V
prove that
<br /> W_1\cap W_2=sp(v2) <br />
its obvious v2 exists in both groups .
how am i supposed to prove it
??
Since
W_1\cap W_2 is a subspace of W1, and W1 has dimension 2, it has dimension 1 or 2. If it has dimension 2, the it is equal to W2. But W_1 contains v1 which is not in W_1\cap W_2 so it is not of dimension 2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top