Let V and W be subspaces of R6 of dimensions 3 and 4 respectively.?

Question

What are the possible dimensions of V intersect W? Explain.

Answer 1

The intersection may have dimension 1, 2, or 3. For examples, let e_i (i = 1, 2, ...,6) denote the standard unit vectors whose i-th component is 1 and all other components are 0.

Example 1. Suppose V is spanned by {e_1, e_2, e_3} and W = span{e_3, e_4, e_5, e_6}. Then V intx W = span{e_1}, so the intersection has dim = 1.

Ex. 2. Take V as above and W = span{e_2, e_3, e_4, e_5}. Then dim(V intx W) = 2.

Ex. 3 Take V as above, W = span{e_1, e_2, e_3, e_4}. Then dim(V intx W) = 3.

Notice that there cannot be an intersection of dim = 4 because the dimension of V is only 3. There cannot be an intersection of dim = 0 because then V intx W = {0}, and V U W would have dim = 7, but V U W is a subset of R^6.

Answer 2

take bases {v_1,v_2,v_3} and {w_1,w_2,w_3,w_4} of V and W, respectively. it is a basic fact of linear algebra that
dim (V+W) + dim (V intersect W) = dim V + dim W
where V+W is the subspace generated by {v_1,v_2,v_3,w_1,w_2,w_3,w_4}.
for a proof of this fact, try this link:
http://planetmath.org/?op=getobj&from=objects&id=8782
now, dim V + dim W = 7 by hypothesis. dim (V+W) can take any of the values 4, 5 and 6: to see this, given the four-dimensional subspace W, we could choose V to be the subspace spanned by {w_1,w_2,w_3} so V+W=W. to get V+W of dimension 5, extend {w_1, ..., w_4} to a basis {w_1, ..., w_4, v_5, v_6} of R^6. then let V be generated by {w_1,w_2, v_5} so V+W is of dimension 5. similarly, if we take V to be the subspace spanned by {w_1, v_5, v_6} then V+W=R^6 is of dimension 6. so the possible values for dim (V intersect W) are 7-6=1, 7-5=2 and 7-4=3.