MA20008 Algebra 1, 2004, Sheet 7 Solutions

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

1. Let V be a two dimensional vector space over ℝ with basis v_{1}, v_{2}. Let W be a two dimensional vector space over ℝ with basis w_{1}, w_{2}. Suppose that α : V ⟶ W is a linear map which has matrix

( \begin{matrix}1 & 2 \\ 3 & 4 \end{matrix} )

with respect to these bases. Determine the matrix of α with respect to new bases v_{1} + 2v_{2}, v_{1} − v_{2} of V and w_{1} + w_{2}, w_{1} − w_{2} of W. Solution Let x_{1} = v_{1}+2v_{2}, x_{2} = v_{1}−v_{2}, and let y_{1} = w_{1}+w_{2}, y_{2} = w_{1} − w_{2}. Now let P be the matrix of id_{V} using the new basis in the domain and the old basis in the codomain. We have id(x_{1}) = x_{1} = v_{1} + 2v_{2} and id(x_{2}) = v_{1} − v_{2}, Therefore

P = ( \begin{matrix}1 & 1 \\ 2 & −1 \end{matrix} ) .

Let Q be the matrix of id_{W} using the old basis in the domain and the new basis in the codomain. We have id(w_{1}) = w_{1} = \frac{1}{2}y_{1} + \frac{1}{2}y_{2} and id(w_{2}) = w_{2} = \frac{1}{2}y_{1} − \frac{1}{2}y_{2}. Therefore

Q = \frac{1}{2} ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ) .

Thus the matrix of α with respect to the new bases is

QAP = ( \begin{matrix}8 & −1 \\ −3 & 0 \end{matrix} ) .

2. Suppose that A is an m × n matrix with entries in a field F. Show that there are vector spaces V and W over F of dimensions n and m respectively, a linear transformation α : V ⟶ W, and choices of bases for these spaces so that the matrix of α with respect to these bases is A. Solution Let V be a vector space over F of dimension n. For example we might have V = F^{n}. Let W be a vector space over F of dimension m. For example we might have V = F^{m}. Choose v_{1}, v_{2}, …, v_{n} a basis of V and w_{1}, w_{2}, …, w_{m} a basis of W. Define a linear map α : V ⟶ W by stipulating that α(v_{j}) = ∑_{i=1}^{m} a_{ij}w_{i} for every j. and extending to all of V by linearity. The matrix of the linear map α is A = (a_{ij}) by design.

3. Let V and W be vector spaces over F of dimensions n and m respectively. Show that there is a choice of bases for V and W such that the matrix of α with respect to these bases is

( \begin{matrix}I_{r} & 0_{r,n−r} \\ 0_{m−r,r} & 0_{m−r,n−r} \end{matrix} ) .

Solution Choose a basis for v_{r+1}, v_{r+2}, …, v_{n} for Ker α and extend it if necessary to a basis v_{1}, v_{2}, …, v_{n} to form a basis of V. Now let w_{i} = α(v_{i}) for 1 ≤ i ≤ r, a collection of linearly independent vectors (q.v. the proof of the rank-nullity theorem). Extend this if necesary to a basis w_{1}, w_{2}, …, w_{m} of W. The matrix of α with respect to these bases has the required shape.

4. Let V = ℝ^{2} viewed as a vector space over ℝ, and thought of as the Euclidean plane. Let ρ_{θ} : V ⟶ V denote the map which is rotation anticlockwise about the origin through θ radians. Choose the standard basis e_{1}, e_{2} for V, where e_{1} = (1, 0) and e_{2} = (0, 1).

(a) Show that ρ_{α} : V ⟶ V is a linear map for every α ∈ ℝ. Solution Perhaps the easiest way to see this is to endow the vector space with a extra structure, as regard V as ℂ, the complex numbers, so that e_{1} is 1 and e_{2} is i. Now anti-clockwise rotation through θ about the origin is accomplished by multiplying by e^{iθ}. Now for w, z ∈ ℂ and λ, μ ∈ ℝ we have

e^{iθ}(λw + μz) = λ(e^{iθ})w + μ(e^{iθ})z

which is exactly what we want.

(b) Show that ρ_{α} ∘ ρ_{β} = ρ_{α+β} for all α, β ∈ ℝ.

Solution This is easy now that we are using ℂ, because this statement amounts to e^{iπ(α+β)} = e^{iπα}e^{iπβ}

(c) Show that ρ_{α} and ρ_{β} commute for all α, β ∈ ℝ. Solution This statement amounts to e^{iπ(α+β)} = e^{iπ(β+α)}.

(d) Write down the matrix of ρ_{α} with respect to the given basis.

Solution

( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) .

(e) Deduce formulas for cos(α + β) and sin(α + β).

( \begin{matrix}cos α & sin α \\ − sin α & cos α \end{matrix} ) ( \begin{matrix}cos β & sin β \\ − sin β & cos β \end{matrix} )

= ( \begin{matrix}cos(α + β) & sin(α + β) \\ − sin(α + β) & cos(α + β) \end{matrix} ) .

Equating the entries yields the standard formulas for cos(α + β) and sin(α + β).

(f) Let c denote complex conjugation. Calculate the matrix of c∘ρ_{θ} ∘c and comment on what this means. Using the standard basis, the matrix of the map defined by complex conjugation is

( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) .

Therefore the required composition is

( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )

= ( \begin{matrix}cos(−θ) & sin(−θ) \\ − sin(−θ) & cos(−θ) \end{matrix} ) .

Thus c ∘ ρ_{θ} ∘ c = ρ_{−θ}.

5. Suppose that A and B are invertible n × n matrices with real entries. Show that there are invertible n × n real matrices D and E such that DAE = B. Pick a basis for V, and use A and this basis to define a linear map α. Now change basis in domain and domain (as in Question 3) so there are n × n invertible matrices P and Q so that QAP = I_{n}. You get the identity matrix because α is bijective. Similarly there are invertible n × n matrices R and S such that RAS = I_{n}. Now R^{−1}P AQS^{−1} = B. Put D = R^{−1}P and E = QS^{−1}. We are done.

6. Define a relation on n × n matrices with real entries as follows: if A, B ∈ Mat_{n,n}(ℝ), we write A ∼ B if and only if there are invertible P, Q ∈ Mat_{n,n}(ℝ) such that QAP = B.

(a) Show that ∼ is an equivalence relation. Solution Let I denote the n × n identity matrix. Now if A ∈ Mat_{n,n}(ℝ), then A = IAI so reflexivity is established. Next suppose that A, B ∈ Mat_{n,n}(ℝ)i and there are invertible P, Q ∈ Mat_{n,n}(ℝ) such that QAP = B. Then Q^{−1}BP^{−1} = A so symmetry is establisheid. Finally suppose that A ∼ B and B ∼ C. Thus there are invertible matrices P_{1}, P_{2}, Q_{1}, Q_{2} ∈ Mat_{n,n}(ℝ) such that P_{1}AQ_{1} = B and P_{2}BQ_{2} = C. Therefore (P_{2}P_{1})A(Q_{1}Q_{2}) = C and both P_{2}P_{1} and Q_{1}Q_{2} are invertible elements of Mat_{n,n}(ℝ). Transitivity is established.

(b) How may equivalence classes are there? Solution There are n + 1 equivalence classes. Every matrix is equivalent to one of the form

( \begin{matrix}I_{r} & 0_{r,n−r} \\ 0_{m−r,r} & 0_{m−r,n−r} \end{matrix} ) .

As explained in lectures, the quantity r is the rank of any linear transformation giving rise to the matrix A, and is both the row rank and column rank of A. Therefore as r varies the matrices

( \begin{matrix}I_{r} & 0_{r,n−r} \\ 0_{m−r,r} & 0_{m−r,n−r} \end{matrix} )

are pairwise inequivalent.