Algebra 1; MA20008; Sheet 3

            G.C.Smith@bath.ac.uk

25-x-2004

1. Suppose that V is a vector space over F, and that S ⊆ V. Let \overline{S} be the intersection of those subspaces of V which contain the subset S, or put formally

\overline{S} = ⋂\{U | U ≤ V, S ⊆ U\}.

Show that \overline{S} = ⟨S⟩. Solution: We have S ⊆ ⟨S⟩ ≤ V. Now ⟨S⟩ is therefore one of the sets being intersected in the definition of \overline{S}. Therefore \overline{S} ⊆ ⟨S⟩. Conversely if S ⊆ U ≤ V and v ∈ ⟨S⟩, then v ∈ U since U is closed under the formation of linear combinations. Therefore ⟨S⟩ ⊆ \overline{S}. Since we have both inclusions it follows that \overline{S} = ⟨S⟩.

2. Let V be a vector space over F and v_{1}, v_{2}, …, v_{n} ∈ V. Suppose that whenever θ_{1}, …, θ_{n}, ψ_{1}, …, ψ_{n} ∈ F and ∑_{i=1}^{n} θ_{i}v_{i} = ∑_{i=i}^{n} ψ_{i}v_{i}, then necessarily λ_{i} = μ_{i} for each i, 1 ≤ i ≤ n. Show that v_{1}, v_{2}, …, v_{n} ∈ V is linearly independent. Solution: It suffices to choose ψ_{i} = 0 for every i, and we obtain the condition for linear independence.

3. Consider V = ℝ as a vector space over ℚ.

(a) Show that 1, \sqrt{2}, \sqrt{3} are linearly independent. Solution: It is first year work to show that \sqrt{2}, \sqrt{3} and \sqrt{6} are irrational, and we assume that you can do this. Suppose that a+b\sqrt{2}+c\sqrt{3} = 0 for rational a, b and c. Therefore (b\sqrt{2}+c\sqrt{3})^{2} ∈ ℚ, so bc\sqrt{6} ∈ ℚ. Now bc = 0 else \sqrt{6} would be rational. If only one of b and c were 0, then \sqrt{2} or \sqrt{3} would be rational, which it isn’t. Therefore b = c = 0, so a = 0 and we are done.

(b) Let α = e^{\frac{πi}{3}} . Which lists of the form 1, α, …, α^{n} are linearly independent? Justify your answer. Solution: α is a root of X^{3} +1 = (X +1)(X^{2} −X +1) but not of X +1 so α is a root of X^{2}−X +1. Thus 1−α+α^{2} = 0 so 1, α, α^{2} is a linearly dependent list, as is 1, α, …, α^{n} whenever n ≥ 2. Also 1 is linearly independent, and 1, α is linearly independent, since a non-trivial linear relation would force α ∈ ℝ, which is false.

(c) Suppose that 1, β, β^{2}, …, β^{n} are linearly independent. Show that 1, (β + 1), (β + 1)^{2}, …, (β + 1)^{n} are linearly independent. Solution: Suppose, for contradiction, that these powers of 1 + β are linearly dependent. Thus there is a non-zero polynomial f (or f(X)) with rational coefficients so that f(1+β) = 0. Now β will be a root of h := f(X +1), and deg h = deg f (and indeed the leading coefficients co-incide). Therefore h is not the zero polynomial and the given powers of β satisfy a non-trivial linear relation. However, we are given that these powers of β are linearly independent over ℚ, so this is absurd. We have the required contradiction.

4. Suppose that X and Y are both linearly independent subsets of V. Does it follow that X ∩ Y is linearly independent? What about X ∪ Y? Solution: A subset of a l.i. set of vectors is l.i. for formal reasons, and X ∩ Y ⊆ X, so we are done. However, the same is not true for the formation of unions. Let V = F = ℝ. Let X = {1}, Y = {2} which are both l.i., but X ∪Y = {1, 2} which is l.d. because 2·1+(−1)·2 = 0.

5. Suppose that V = U ⊕ W. We are given a sets of vectors X ⊆ U and Y ⊆ W. Is X ∪ Y necessarily a linearly independent set of vectors? Solution: We have proved that a direct sum yields uniqueness of decomposition, so if v ∈ V and v = u_{1} + w_{1} = u_{2} + w_{2} for u_{1}, u_{2} ∈ U and w_{1}, w_{2} ∈ W, then u_{1} = u_{2} and w_{1} = w_{2}. Now suppose that we have scalars λ_{i}, θ_{j} so that

∑_{i=1}^{m} λ_{i}u_{i} + ∑_{j=1}^{n} θ_{j}w_{j} = 0.

Here u_{1}, …, u_{m} ∈ X and v_{1}, …, v_{n} ∈ Y. The uniqueness of expression, compared to 0 + 0 = 0, ensures that both

∑_{i=1}^{m} λ_{i}u_{i} = 0

and

∑_{i=1}^{n} θ_{j}w_{j} = 0.

The l.i. of both X and Y forces all scalars to vanish, and we are done.

6. Suppose that v_{1}, v_{2}, …, v_{n} is a linearly independent list of vectors in the vector space V. We are given w ∈ V. Does it follow that

v_{1} + w, v_{2} + w, …, v_{n} + w

are linearly independent? Solution: No. Choose w = −v_{1} and the zero vector occurs in the list.

7. Suppose that v_{1}, v_{2}, …, v_{n} is a linearly dependent list of vectors in the vector space V. We are given w ∈ V. Does it follow that

v_{1} + w, v_{2} + w, …, v_{n} + w

is linearly dependent? Solution: No. Let V = ℝ, F = ℝ. Let n = 1 and v_{1} = 0. Let w = 1.

8. Let V = ℝ^{3} viewed as vector space over ℝ. Let v_{1}, …, v_{8} be the position vectors of the vertices of a cube.

(a) Let

A = {∑_{i} λ_{i}v_{i} | 0 ≤ λ_{i} ≤ 1 for all i, ∑_{i} λ_{i} = 1} .

Describe the set A, viewed as a collection of position vectors, geometrically. Solution: The given position vectors point to the points inside and on the surface of the cube.

(b) Let

B = {∑_{i} λ_{i}v_{i} | λ_{i} ≥ 0 for all i, } .

Under what circumstances is B = ℝ^{3}? Under what circumstances is B a closed half space (i.e. one side of a plane and all the points on that plane)? What other shapes can arise? Solution: We have B = ℝ^{3} exactly when the origin is strictly inside the cube. If the origin is in the interior of a face, then B is a half-space. If the origin is in the interior of an edge, then B is the intersection of two half-spaces defined by perpendicular planes. If the origin is at a vertex of the cube, then B is the intersection of three half-spaces defined by pairwise perpendicular planes (also known as an octant). If the origin is strictly outside the cube, then B will be an infinite cone with finitely many planar faces.