Here is a linear algebra question :?

Question

Consider the following subspaces of F^5 (where F is a field):

W1= { (a1, a2, a3, a4, a5) : a1=a2 + a4, a3=a5}, and
W2= { (a1, a2, a3, a4, a5) : a1 = a3 = a4}.
1. Describe the intersection of W1 and W2 and find a basis for this subspace.
2. Use the dimension formula to show that W1 + W2 = F^5
3. Starintg with the basis in (1) for the intersection of W1 and W2, find bases for W1, W2, and use them to find a basis for F^5.
Please give reasons.

Answer 1

#1: W₁∩W₂ = {(a₁, a₂, a₃, a₄, a₅): a₁ = a₃ = a₄ = a₅, a₂=0}. It can be generated by the basis {(1, 0, 1, 1, 1)}.

Proof: We know that if a = (a₁, a₂, a₃, a₄, a₅)∈W₁∩W₂, then a₁ = a₃ = a₄ (since it is in W₂), and a₃=a₅ (since it is in W₁). This means a₁ = a₃ = a₄ = a₅. We also know that a∈W₁∩W₂ implies that a₁ = a₂+a₄, but since a₁ = a₄, this means a₂=0. Conversely, it is easy to see that if a₁ = a₃ = a₄ = a₅ and a₂=0, then a∈W₁∩W₂, so W₁∩W₂ is precisely the set of vectors for which the conditions above are true. From this it follows that every vector in W₁∩W₂ is given by a₁(1, 0, 1, 1, 1), so {(1, 0, 1, 1, 1)} is a basis for that subspace.

#2: We know that W₁ is of dimension not less than 3, since (1, 1, 0, 0, 0), (1, 0, 0, 1, 0), and (0, 0, 1, 0, 1) are three linearly independent vectors in W₁. We also know that W₂ is of dimension not less than 3, since (1, 0, 1, 1, 0), (0, 1, 0, 0, 0), and (0, 0, 0, 0, 1) are three linearly independent vectors in W₂. Thus, dim (W₁ + W₂) = dim W₁ + dim W₂ - dim (W₁∩W₂) ≥ 3 + 3 - 1 = 5, but dim F⁵ = 5, so W₁ + W₂ must in fact be all of F⁵ (and in fact dim W₁ = 3 and dim W₂ = 3, for if either one was greater, then dim (W₁+W₂) would actually be greater than 5, which is absurd. This also implies that the sets of vectors we gave above are bases of W₁ and W₂, respectively).

3: {(1, 0, 1, 1, 1), (1, 1, 0, 0, 0), (1, 0, 0, 1, 0), (0, 1, 0, 0, 0), (0, 0, 0, 0, 1)} is a basis for F⁵, constructed from the bases for W₁ and W₂.

Answer 2

There are no prerequisites really: linear algebra is pretty self contained.

Answer 3

Sam, you already have a good answer. Go with it.