## Intersection of subspaces

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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 Recognitions: Homework Help Science Advisor 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."