What Is the Intersection of Subspaces U and V in R^3?

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SUMMARY

The intersection of the subspaces U and V in R^3 is not the entirety of R^3, as both subspaces are defined with specific restrictions. Subspace U is defined by the equations a1 = 3(a2) and a3 = -a2, while subspace V is defined by the equation a1 - 4(a2) - a3 = 0. By substituting the equations from U into V, it is confirmed that the intersection is indeed a non-zero point, indicating that U and V are equivalent (U = V). Therefore, the intersection is a one-dimensional subspace rather than the whole R^3.

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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators looking to deepen their understanding of vector spaces and their intersections.

loli12
I have 2 subspaces U and V of R^3 which
U = {(a1, a2, a3) in R^3: a1 = 3(a2) and a3 = -a2}
V = {(a1, a2, a3) in R^3: a1 - 4(a2) - a3 = 0}

I used the information in U and substituted it into the equation in V and I got 0 = 0. So, does it mean that the intersection of U and V is the whole R^3 which has no restrictions on a1, a2 and a3 (they are free)? Or do the original restrictions on both the original subspaces still being applied to the intersection?
 
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The intersection of U and V cannot possibly be all of R³. How could the intersection of two sets be bigger than both of the sets? Both subspaces are 1-dimensional, so the intersection is either 1-dimensional or 0-dimensional. Can you find a non-zero point that is in both U and V? If so, then the intersection of U and V is U and is also V (i.e. U = V). A point in U takes the form (x, x/3, -x/3). Would such a point be in V?

x - 4(x/3) - (-x/3) = x - (4/3)x + (1/3)x = 0

so the answer is "yes."
 

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