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

[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?

 

[AKG]

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."

 

[quicknote]

The intersection of two subspaces, U and V, is defined as the set of all vectors that are in both U and V. In this case, the intersection of U and V would be the set of all vectors that satisfy the conditions of both U and V.

Using the given information, we can see that the conditions for U and V are not contradictory, as substituting the conditions of U into the equation of V results in 0=0. This means that all vectors in U are also in V, and vice versa. Therefore, the intersection of U and V is the set of all vectors in R^3 that satisfy the conditions of both U and V.

In terms of the restrictions, the original restrictions for both U and V are still being applied to the intersection. This means that any vector in the intersection must satisfy the conditions of both U and V, which are the restrictions given in the problem. So, the intersection is not the whole R^3, but rather a subset of R^3 that satisfies both sets of restrictions.