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.
Re: 2 Linear Algebra Proofs dealing with vector spaces and subspaces
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.