MA20008 Algebra 1, 2004, Sheet 6 Solutions

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

1. Let V be a vector space of dimension n. Suppose that α : V → V is a linear map. Show that the following are equivalent.

(a) α is injective.

(b) α is bijective.

(c) α is surjective.

(d) There is a basis v_{1}, …, v_{n} of V such that α(v_{1}), …, α(v_{n}) is a basis of V.

(e) For every basis v_{1}, …, v_{n} of V, the vectors α(v_{1}), …, α(v_{n}) also form a basis of V.

Solution The rank nullity theorem tells us that α is injective if and only if it is surjective. Therefore (a), (b) and (c) are equivalent. Now (c) implies (e) since a spanning set must map to a spanning set, and a spanning set of size n must be a basis. Certainly (e) implies (d). However, condition (d) forces the image of α to include a spanning set, and so (c) is forced. We are done.

2. Suppose that

X = ( \begin{matrix}A & B \\ 0 & C \end{matrix} )

is a 2r by 2r matrix built from the four r by r matrices A, B, C and the zero matrix 0. Suppose that X has an inverse matrix. Describe that matrix in terms of A, B, C and 0. Solution Let the inverse matrix be of the shape

Y = ( \begin{matrix}D & E \\ F & G \end{matrix} )

where D, E, F and G are r by r matrices. Now (a) AD + BF = I_{r},

(b) AE + BG = 0,

(c) CF = 0 and

(d) CG = I_{r}. From (d) G = C^{−1}. From (c) and premultiplication by C^{−1} we obtain F = 0. Next (a) forces D = A^{−1} an d (b) implies that AE + BC^{−1} = 0 so E = −A^{−1}BC^{−1}.

3. The matrix

F = ( \begin{matrix}1 & 1 \\ 1 & 0 \end{matrix} )

has entries in the Field F_{7}, the integers modulo 7. Calculate

(a) F^{2}.

(b) F^{5}.

(c) F^{1000}. (d)

∑_{i=0}^{999} F^{i}

where F^{0} denotes the identity matrix. Solution For integers n ≥ 0 define the Fibonacci sequence mod 7 via F_{0} = 0 and F_{1} = 1 where F_{0}, F_{1} ∈ ℤ_{7}. Then inductively F_{n} = F_{n−1} + F_{n−2} whenever n ≥ 2. Thus

(F_{n}) = 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, ….

Thus F_{16} = 0 and F_{17} = 1 so (F_{n}) is periodic with fundamental period 16. Now

F^{n} = ( \begin{matrix}F_{n+1} & F_{n} \\ F_{n} & F_{n−1} \end{matrix} )

by induction on n. Therefore (a)

F^{2} = ( \begin{matrix}F_{3} & F_{2} \\ F_{2} & F_{1} \end{matrix} ) = ( \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ) .

(b)

F^{5} = ( \begin{matrix}F_{6} & F_{5} \\ F_{5} & F_{4} \end{matrix} ) = ( \begin{matrix}1 & 5 \\ 5 & 3 \end{matrix} ) .

(c)

F^{1000} = ( \begin{matrix}F_{1001} & F_{1000} \\ F_{1000} & F_{999} \end{matrix} ) = ( \begin{matrix}F_{9} & F_{8} \\ F_{8} & F_{7} \end{matrix} ) = ( \begin{matrix}6 & 0 \\ 0 & 6 \end{matrix} ) .

(d) The matrix

F = ( \begin{matrix}1 & 1 \\ 1 & 0 \end{matrix} )

has entries in ℚ, Let I denote the 2 by 2 identity matrix. Show that I and F are linearly independent but that I, F and F^{2} are linearly dependent elements of the vector space of 2 by 2 matrices with rational entries (with scalars in ℚ). Solution By inspection or from the Fibonacci recurrence F^{2} = F + 1 so F^{2} − F − 1 is a non-trivial linear relation among 1, F and F^{2}. Suppose that λ, μ ∈ ℚ and λI + μF = 0 then

i. λ + μ = 0, ii. μ = 0, iii. λ + μ = 0 and iv. 0 + 0 = 0.

by inspecting the four positions in turn. These equations force λ = μ = 0 as required.

4. Suppose that α, β : V ⟶ V are a pair of commuting linear maps.

(a) Prove that both Im α and Ker α are β-invariant spaces. Solution Suppose that v ∈ Im α so there is u ∈ V such that α(u) = v. Now β(v) = β(α(u)) = α(β(u)) ∈ Im α. ON the other hand, suppose that x ∈ Ker α, then α(β(x)) = β(α(x)) = β(0) = 0. Thus β(x) ∈ Ker α as required.

(b) Prove that Im α + Im β is both α-invariant and β-invariant. Solution Each of Im α and Im β is both α-invariant and βinvariant. It follows that their sum has the same property, because if x ∈ Im α and y ∈ Im β, then α(x + y) = α(x) + α(y) which is the sum of two vectors, the first of which is in Im α and the second in Im β. The sum of subspaces is also β-invariant by symmetry.

(c) Prove that Im α ∩ Im β is both α-invariant and β-invariant. Solution This is a formality. If x is in the intersection, then α(x) ∈ Im α and α(x) ∈ Im β so α(x) ∈ Im α ∩ Im β. The same holds for β by symmetry.