Thread: 2 Linear Algebra Proofs dealing with vector spaces and subspaces

1. Let V be the set of all pairs (x; y) of real numbers and suppose vector
addition and scalar multiplication are defined in the following way:
(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)
a(x, y) = (ax, y),
for any scalar a in the field of real numbers. Is the set V , with these operations,
a vector space over the field of real numbers? Justify your answer.

2. Let W1 and W2 be subspaces of a vector space V such that their union W1 U W2 (W1 U W2 is the set of vectors which belong to either W1 or W2) is also
a subspace of V . Prove that either W1 is contained in W2 or vice versa.

Thank you!

1):

consider a = 1, b = 1, x = y = 1.

isn't it an axiom that (a + b)(x,y) = a(x,y) + b(x,y)? what does the above example lead you to conclude?

2): if W₁ ⊆ W₂, there is nothing to prove, so we lose no generality by assuming this is NOT the case. thus there is some w₁ in W₁ that is NOT in W₂.

so let w₂ be any element of W₂. consider w = w₁+w₂. since w₁ is in W₁, and w₂ lies in W₂, each term lies in W₁UW₂.

since W₁UW₂, is a subspace, it is closed under addition hence w = w₁+w₂ is in W₁UW₂.

we thus have:

w₁ = w - w₂.

since this is NOT in W₂, w cannot be in W₂ either, or else w - w₂ = w + (-1)w₂ would be in W₂, since W₂ is a subspace (and closed under vector addition and scalar multiplication).

but since w is in W₁UW₂, if it is not in W₂, it must lie in W₁.

so w₂ = w - w₁ lies in W₁ (since W₁ is a subspace), and since w₂ was chosen arbitrarily from W₂, we have shown if W₁ is not a subset of W₂, then W₂ is a subset of W₁.