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!

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

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 W_{1} ⊆ W_{2}, there is nothing to prove, so we lose no generality by assuming this is NOT the case. thus there is some w_{1} in W_{1} that is NOT in W_{2}.

so let w_{2} be any element of W_{2}. consider w = w_{1}+w_{2}. since w_{1} is in W_{1}, and w_{2} lies in W_{2}, each term lies in W_{1}UW_{2}.

since W_{1}UW_{2}, is a subspace, it is closed under addition hence w = w_{1}+w_{2} is in W_{1}UW_{2}.

we thus have:

w_{1} = w - w_{2}.

since this is NOT in W_{2}, w cannot be in W_{2} either, or else w - w_{2} = w + (-1)w_{2} would be in W_{2}, since W_{2} is a subspace (and closed under vector addition and scalar multiplication).

but since w is in W_{1}UW_{2}, if it is not in W_{2}, it must lie in W_{1}.

so w_{2} = w - w_{1} lies in W_{1} (since W_{1} is a subspace), and since w_{2} was chosen arbitrarily from W_{2}, we have shown if W_{1} is not a subset of W_{2}, then W_{2} is a subset of W_{1}.