Course MT3818 Topics in Geometry

Solutions 7

  1. λ1a1 = a1 - λ2a1 - λ3a1 ... - λra1 and so λ1a1 + λ2a2 + ... + λrar = a1 + λ2(a2 - a1) + λ3(a3 - a1) + ... + λr(ar - a1) and so the span is the translation by a1 of the linear subspace spanned by { (a2 - a1), (a3 - a1), ... , (ar - a1) }.

  2. Suppose we have μ2(a2 - a1) + μ3(a3 - a1) + ... + μr(ar - a1) = 0. Then -(μ2 + μ3 + ... + μr)a1 + μ2a2 + ... + μrar = 0 and since the sum of the coefficients is 0 it follows that μ2 = μ3 = ... = μr = 0 and so the given set is linearly independent.

    To prove that (a1, 1), (a2, 1), ... , (ar , 1) are linearly independent, suppose that λ1(a1, 1) + λ2(a2, 1) + ... + λr(ar , 1) = 0 in Rn+1. Then we have λ1a1 + λ2a2 + ... + λrar = 0 in Rn and λ1 + λ2 + ... + λr = 0 in R and so from the definition of affine independence we have λ1 = λ2 = ... = λr = 0 and so the given set is linearly independent.

    Given two affinely independent sets { a0 , a1 , ... , an } and { b0 , b1 , ... , bn } in Rnthe sets { a1 - a0 , ... , an - a0 } and { b1 - b0 , ... , bn - b0 } are linearly independent and so there is a unique linear map L taking one to the other. Then the map Tb0 ∘ L ∘T-a0 is the required affine map. It is easily seen to be unique.

  3. As in the Question 3 of Exercises 3, let M be the matrix with A ∈GL(n, R). Then M maps a column to a similar column and so acts on this affine subspace by ↦ where y = Ax + (a1, ... , an) Hence M acts as an affine transformation.
    As in the last example, the n + 1 affinely independent points give n + 1 linearly independent points whose image determines a matrix and hence the map.

  4. Suppose f is Ta ∘ L with L linear. Then f(λ1a1 + λ2a2 + ... + λrar) = a + λ1L(a1) + λ2L(a2) + ... + λrL(ar).
    λ1f(a1) + λ2f(a2) + ... + λrf(ar) = λ1(a + L(a1)) + λ2(a + L(a2)) + ... + λr(a + L(ar)) which is the same since λ1 + λ2 + ... + λr = 0.
    The centroid of { a1 , a2 , ... , ar } is 1/n(a1 + a2 + ... + ar) and so this is mapped to 1/n(f(a1) + f(a2) + ... + f(ar)) as required.

  5. A line joining the midpoints of two sides of a triangle is parallel to the third side.


    Hence the pair of sides shown is parallel. Similarly for the other pair of sides.


    The centroid of the quadrilateral is 1/4 (a + b + c + d). This is also the centroid of the parallelogram: 1/4 [(a + b)/2 + (b + c)/2 + (c + d)/2 + (d + a)/2] whose centroid is the meet of its diagonals.
    This result does hold in R3 also.

  6. To find a fixed point, we solve (a + L ∘ λI)x = x. That is (I - λL)x = a with L ∈ O(2). But L can only have eigenvalues of ±1 and so the transformation I - λL is invertible unless λ = ±1 (when the transformation will be an isometry). Hence we can solve the equation to find a unique fixed point.

  7. Take a square ABCD on AB and two possible squares A'B'C'D' and A'B'C''D'' on A'B'.
    Note that similarity transformations map squares to squares.
    There is a unique affine map taking ΔABC to ΔA'B'C' and another (opposite) affine map taking ΔABC to ΔA'B'C'' and these are the required pair of similarities.