MA20008 Algebra 1, 2004, Sheet 5

Geoff Smith, http://www.bath.ac.uk/∼masgcs

1. Let V be a vector space of dimension n.

(a) Suppose that we have subspaces

0 = V_{0} < V_{1} < V_{2} ⋯ < V_{n} = V

with dim V_{i} = i for every i = 0, 1, …, n. For each i > 0 choose v_{i} ∈ V_{i} but v_{i} ∉ V_{i−1}. Show that

v_{1}, v_{2}, …, v_{n}

is a basis of V. Solution It suffices to demonstrate linear independence. Suppose, for contradiction, that

∑_{i=1}^{r} λ_{i}v_{i}

with λ_{r} ≠ 0. Then

v_{r} = −λ_{r}^{−1} (∑_{i=1}^{r−1} λ_{i}v_{i})

so v_{r} ∈ V_{r−1}, a contradiction.

(b) Suppose that

v_{1}, v_{2}, …, v_{n}

is a basis of V. Show that there are subspaces

0 = V_{0} < V_{1} < V_{2} ⋯ < V_{n} = V

with dim V_{i} = i for every i = 0, 1, …, n such that v_{i} ∈ V_{i} but v_{i} ∉ V_{i−1} for every i > 0. Solution Let V_{i} = ⟨v_{1}, v_{2}, …, v_{i}⟩. The linear independence of these vectors ensures that dim V_{i} = i.

2. Let V be a vector space of dimension n. Suppose that α : V ⟶ V is a linear map.

(a) Show that V ≥ Im α ≥ Im α^{2} ≥ … ≥ 0. Observation: Im α^{r} denotes the map defined by composing r copies of α. Solution Suppose that v ∈ Im α^{r+1}, so there is w ∈ V such that v = α^{r+1}(w) = α^{r}(α((w)) ∈ Im α^{r}.

(b) Show that 0 ≤ Ker α ≤ Ker α^{2} ≤ ⋯ ≤ V.

Solution Suppose that v ∈ Ker α^{r}, then α^{r}(v) = 0 so α^{r+1}(v) = α(0) = 0. Therefore v ∈ Ker α^{r+1}.

(c) Suppose that r is a natural number and Im α^{r} = Im α^{r+1}. Show that Im α^{t} = Im α^{r} for all natural numbers t ≥ r. Solution For all v ∈ V we have α^{r}(v) = α^{r+1}(v_{1}) for some v_{1} ∈ V. Now α^{r+1}(v_{1}) = α(α^{r}(v_{1})) = α(α^{r+1}(v_{2})) for some v_{2} ∈ V. Continuing in this way we have α^{r}(v) = α^{r+i}(v_{i}) for some v_{i} ∈ V for every positive i. Therefore Im α^{r} ≤ Im α^{r+i} for all positive i. However we established that Im α^{r+i} ≤ Im α^{r} in 2(a). We are done.

(d) Let V be a vector space of dimension n. Suppose that α : V ⟶ V is a linear map such that α^{2} = 0 (the zero map). Show that the nullity ν_{α} satisfies 2ν_{α} ≥ n. Solution Im α ≤ Ker α so ρ_{α} ≤ ν_{α}. The rank-nullity theorem asserts that ρ_{α} + ν_{α} = n. Therefore ν_{α} + ν_{α} ≥ n and we are done.

(e) Suppose that there is a natural number m such that α^{m} is the zero map 0. Prove that α^{n} is 0. Solution This follows from parts (a) and (c) and observing that the dimension of Im α^{i} is no more that n − i.

3. Suppose that V is a vector space of dimension n and that α : V ⟶ V is a linear map. Suppose that α^{3} is 0, the zero map. Show that ν_{α} ≥ n/3 where ν_{α} denotes the nullity of α. Hint: let V_{0} = V, V_{1} = Im α and V_{2} = Im α^{2}. Let α_{1} : V ⟶ V_{1} and α_{2} : V_{1} ⟶ V_{2} be the maps defined by α_{i}(v) = α(v) for all v ∈ V_{i−1}. Deploy the rank nullity theorem. Solution Let ρ, ν denote the rank and nullity of α. Also let ρ_{i}, ν_{i} denote the rank and nullity of α_{i}. Now n = ν_{1} + ρ_{1}, ρ_{1} = ν_{2} + ρ_{2}. Also ν_{1} = ν and ν_{2} ≤ ν. Moreover ρ_{2} ≤ ν so 3ν ≥ n and we are done.

4. Let V be a finite dimensional vector space, with W, X, Y and Z all subspaces of V.

(a) Show that

dim(X ∩ Y) + dim((X + Y) ∩ Z)
= dim(Y ∩ Z) + dim((Y + Z) ∩ X)
= dim(Z ∩ X) + dim((Z + X) ∩ Y).

Solution dim(X + Y + Z) = dim((X + Y) + Z) = dim(X + Y) + dim Z − dim((X + Y) ∩ Z) = dim X + dim Y + dim Z − dim(X ∩ Y) − dim((X + Y) ∩ Z). Permute the subspaces and we are done.

(b) Show that

dim(W ∩ X) + dim(Y ∩ Z) + dim((W + X) ∩ (Y + Z))

= dim(W ∩ Y) + dim(X ∩ Z) + dim((W + Y) ∩ (X + Z))
= dim(W ∩ Z) + dim(X ∩ Y) + dim((W + Z) ∩ (X + Y))

Solution W + X + Y + Z = (W + X) + (Y + Z) = (W + Y) + (X + Z) = (W + Z) + (X + Y). Take dimensions and you are done.

Hint: this question is not hard. You need an idea. Please do not start picking bases; that way madness lies.