Course MT3818 Topics in Geometry

Solutions 7

  1. lambda1a1 = a1 - lambda2a1 - lambda3a1 ... - lambdara1 and so lambda1a1 + lambda2a2 + ... + lambdarar = a1 + lambda2(a2 - a1) + lambda3(a3 - a1) + ... + lambdar(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 mu2(a2 - a1) + mu3(a3 - a1) + ... + mur(ar - a1) = 0. Then -(mu2 + mu3 + ... + mur)a1 + mu2a2 + ... + murar = 0 and since the sum of the coefficients is 0 it follows that mu2 = mu3 = ... = mur = 0 and so the given set is linearly independent.

    To prove that (a1, 1), (a2, 1), ... , (ar , 1) are linearly independent, suppose that lambda1(a1, 1) + lambda2(a2, 1) + ... + lambdar(ar , 1) = 0 in Rn+1. Then we have lambda1a1 + lambda2a2 + ... + lambdarar = 0 in Rn and lambda1 + lambda2 + ... + lambdar = 0 in R and so from the definition of affine independence we have lambda1 = lambda2 = ... = lambdar = 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 comp L compT-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 belongsGL(n, R). Then M maps a column to a similar column and so acts on this affine subspace by goesto 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 comp L with L linear. Then f(lambda1a1 + lambda2a2 + ... + lambdarar) = a + lambda1L(a1) + lambda2L(a2) + ... + lambdarL(ar).
    lambda1f(a1) + lambda2f(a2) + ... + lambdarf(ar) = lambda1(a + L(a1)) + lambda2(a + L(a2)) + ... + lambdar(a + L(ar)) which is the same since lambda1 + lambda2 + ... + lambdar = 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 comp lambdaI)x = x. That is (I - lambdaL)x = a with L belongs O(2). But L can only have eigenvalues of plusminus1 and so the transformation I - lambdaL is invertible unless lambda = plusminus1 (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 bigdeltaABC to bigdeltaA'B'C' and another (opposite) affine map taking bigdeltaABC to bigdeltaA'B'C'' and these are the required pair of similarities.