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

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The discussion focuses on proving that the intersection of two subspaces, W_1 and W_2, defined as W_1=sp(v1,v2) and W_2=sp(v2,v3) in a vector space V over field F, is equal to sp(v2). It is established that since W_1 has dimension 2, the intersection W_1∩W_2 must have a dimension of either 1 or 2. However, since W_1 contains the vector v1, which is not present in the intersection, the dimension cannot be 2, confirming that W_1∩W_2=sp(v2).

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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
??
 
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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.
 

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