Algebra 1; MA20008; Sheet 4 Solutions

            G.C.Smith@bath.ac.uk

2-xi-2004

1. Suppose that n is a natural number or 0, and F is a field. Show that there is a vector space over F of dimension n. Solution The zero space has basis ∅ and so has dimension |∅| = 0. If n > 0, then F^{n} has dimension n.

2. Suppose that V is a vector space with subspaces U and W both of dimension n < ∞. Does it follow that V is finite dimensional? Does it follow that V has dimension n? Does it follow that U = W? In each case you should supply a reason for your answer. Solution The answers are no, no and no. To illustrate all three points, let F be a field and let V = F [X] be the set of polynomials in X with coefficients in F. This V is not a vector space of finite dimension. Let U be the subset of V consisting of polynomials of degree at most 1. Let W be the subset of V consisting of polynomials of degree at most 2 but which have constant term 0. Now U ≠ W, dim U = dim W = 2 but dim V ≠ 2.

3. Let V be a vector space of dimension n. Suppose that V_{0}, V_{1}, …, V_{m} are subspaces of V with

V_{0} ≤ V_{1} ≤ ⋯ ≤ V_{m}.

(a) Suppose that m > n. Show that there is i ∈ {1, 2, …, m} such that V_{i} = V_{i−1}. Solution The dimensions of the spaces V_{0}, V_{1}, …, V_{m} are weakly increasing. If m > n, then two spaces V_{i} and V_{i−1} must have the same dimension, and therefore (theorem in lectures) V_{i−1} = V_{i}.

(b) Suppose that m ≤ n. Show that it may be that the spaces

V_{0}, V_{1}, …, V_{m}

are distinct. Solution Let v_{1}, v_{2}, …, v_{n}. Let V_{j} be the span of v_{1}, v_{2}, …, v_{j}. This does the job.

4. Suppose that α : U ⟶ W is a linear map between vector spaces over the same field. Let x_{1}, x_{2}, …, x_{n} be vectors in U.

(a) Suppose that U = ⟨x_{1}, x_{2}, …, x_{n}⟩ and α is surjective. Prove that W = ⟨α(x_{1}), α(x_{2}), …, α(x_{n})⟩. Solution Suppose that w ∈ W. Since α is surjective there is u ∈ U sich that α(u) = w. Now u = ∑_{i} λx_{i} so w = α (∑_{i} λ_{i}x_{i}) = ∑_{i} λ_{i}α(x_{i}).

(b) Suppose that x_{1}, x_{2}, …, x_{n} are linearly independent and α is injective. Show that α(x_{1}), α(x_{2}), …, α(x_{n}) are linearly independent. Solution Suppose that ∑_{i} λ_{i}α(x_{i}) = 0. Therefore α(∑_{i} λ_{i}x_{i}) = 0. Now by injectivity ∑_{i} λ_{i}x_{i} = 0. However, the x_{i} are linearly independent, so each λ_{i} is 0.

5. Let ζ = e^{\frac{2πi}{5}} ∈ ℂ.

(a) Suppose that we view ℂ as a vector space over ℚ. Show that 1, ζ, ζ^{2}, ζ^{3} are linearly independent. Solution ζ is a root of X^{4} + X^{3} + X^{2} + X + 1. However, ζ is not a root of any rational polynomial of smaller degree. To see this, note that a smallest degree non-zero rational polynomial having α as a root must divide X^{4} + X^{3} + X^{2} + X + 1. First eliminate the possibility of a linear factor, and then a quadratic factor.

(b) Suppose that we view ℂ as a vector space over ℝ. Show that 1, ζ, ζ^{2}, ζ^{3} are linearly dependent. Solution \overline{ζ} = ζ^{4} = ζ^{−1}. Now (X − ζ)(X − \overline{ζ}) is a real polynomial of degree 2 which has ζ as a root.

6. Let V be a vector space with subspaces U, W such that U and W are both finite dimensional. Let u_{1}, u_{2}, …, u_{m} be a basis of U and w_{1}, w_{2}, …, w_{n} be a basis of W.

(a) Show that U + W is finite dimensional. Solution Certainly U + W = ⟨u_{1}, u_{2}, …, u_{m}, w_{1}, w_{2}, …, w_{n}⟩, so U + W has a finite spanning set and therefore a finite basis.

(b) Show that u_{1}, u_{2}, …, u_{m}, w_{1}, w_{2}, …, w_{n} need not be a basis of U + W. Solution Well, perhaps u_{1} = w_{1}. That would do.

(c) Suppose that U + W = U ⊕ W. Show that

u_{1}, u_{2}, …, u_{m}, w_{1}, w_{2}, …, w_{n}

is a basis of U + W. Solution Spanning is not an issue; linear independence is. Suppose that there are scalars λ_{i}, μ_{i} such that σ_{i}(λ_{i}u_{i})+σ_{j}(μ_{j}w_{j}) = 0. Since U +W = U ⊕W it follows that σ_{i}(λ_{i}u_{i}) = 0 and σ_{i}μ_{j}w_{j} = 0. Now λ_{i} and μ_{j} are 0 for every i and j by the linear independence of the relevant sequences.

(d) Suppose that u_{1}, u_{2}, …, u_{m}, w_{1}, w_{2}, …, w_{n} is a basis of U + W. Show that U + W = U ⊕ W. Solution It suffices to show that U ∩ W = 0. Suppose not, then there is v ∈ U ∩W with v ≠ 0. Now v ∈ U so v = ∑ θ_{i}u_{i} and v = ∑ ψ_{j}u_{j}. This yields a non-trivial linear relation ∑ θ_{i}u_{i} − ∑ ψ_{j}w_{j} among the vectors u_{i} and w_{i}, which cannot therefore be linearly independent.

7. Let I be a set. Let V be the set of real valued functions on I; more

formally

V = {f|f : I ⟶ ℝ} .

Define addition on V by (f + h)(x) := f(x) + g(x) for all x ∈ I. If λ ∈ ℝ and f ∈ V we define λ · f ∈ V by (λ · f)(x) = (λ)(f(x)) where the final multiplication is just the product (in ℝ).

(a) Check that V is now a vector space over ℝ. Solution This is routine.

(b) For each i ∈ I, define a function δ_{i} ∈ V where δ_{i}(x) = δ_{i,x} (Krönecker delta). Thus δ_{i}(i) = 1 and δ_{i}(x) = 0 if x ≠ i. Show that the vectors δ_{i} are linearly independent. Solution Suppose that λ_{i} are scalars (all but finitely many of which are 0). Then ∑′_{i} λ_{i}δ_{i} = 0. Choose any j ∈ I¡ then λ_{i} are scalars (all but finitely many of which are 0). Then ∑′_{i} λ_{i}δ_{i})(j) = 0 so λ_{j} = 0. Thus these maps are linearly independent.

(c) Let W = ⟨δ_{i} : i ∈ I⟩ be the span of all the δ_{i}. Show that the vectors δ_{i} form a basis of W (in that they are a linearly independent spanning set for W). Well they are linearly independent by the previous answer, and course they span W by the design of W.

(d) Show that W = V if and only if I is finite. W is the subset of V consisting of functions of finite support, i.e. functions which take non-zero values at only finitely many elements of the domain. The two subsets co-incide if and only if I is finite.

(e) Give an explicit example of a vector space with an clearly describable uncountable basis (no set theoretic metaphysics allowed). Solution See part (c).