Linear Algebra/Topic: Orthonormal Matrices/Solutions

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Problem 1

Decide if each of these is an orthonormal matrix.

  1. \begin{pmatrix}
1/\sqrt{2} &-1/\sqrt{2}  \\
-1/\sqrt{2} &-1/\sqrt{2}
  2. \begin{pmatrix}
1/\sqrt{3} &-1/\sqrt{3}  \\
-1/\sqrt{3} &-1/\sqrt{3}
  3. \begin{pmatrix}
1/\sqrt{3} &-\sqrt{2}/\sqrt{3}  \\
-\sqrt{2}/\sqrt{3} &-1/\sqrt{3}
  1. Yes.
  2. No, the columns do not have length one.
  3. Yes.
Problem 2

Write down the formula for each of these distance-preserving maps.

  1. the map that rotates \pi/6 radians, and then translates by \vec{e}_2
  2. the map that reflects about the line y=2x
  3. the map that reflects about y=-2x and translates over 1 and up 1

Some of these are nonlinear, because they involve a nontrivial translation.

  1. \begin{pmatrix} x \\ y \end{pmatrix}
x\cdot\cos(\pi/6)-y\cdot\sin(\pi/6) \\
+\begin{pmatrix} 0 \\ 1 \end{pmatrix}
x\cdot(\sqrt{3}/2)-y\cdot(1/2)+0 \\
  2. The line y=2x makes an angle of \arctan(2/1) with the x-axis. Thus \sin\theta=2/\sqrt{5} and \cos\theta=1/\sqrt{5}.
\begin{pmatrix} x \\ y \end{pmatrix}
x\cdot(1/\sqrt{5})-y\cdot(2/\sqrt{5}) \\
  3. \begin{pmatrix} x \\ y \end{pmatrix}
x\cdot(1/\sqrt{5})-y\cdot(-2/\sqrt{5}) \\
+\begin{pmatrix} 1 \\ 1 \end{pmatrix}
x/\sqrt{5}+2y/\sqrt{5}+1 \\
Problem 3
  1. The proof that a map that is distance-preserving and sends the zero vector to itself incidentally shows that such a map is one-to-one and onto (the point in the domain determined by d_0, d_1, and d_2 corresponds to the point in the codomain determined by those three). Therefore any distance-preserving map has an inverse. Show that the inverse is also distance-preserving.
  2. Prove that congruence is an equivalence relation between plane figures.
  1. Let f be distance-preserving and consider f^{-1}. Any two points in the codomain can be written as f(P_1) and f(P_2). Because f is distance-preserving, the distance from f(P_1) to f(P_2) equals the distance from P_1 to P_2. But this is exactly what is required for f^{-1} to be distance-preserving.
  2. Any plane figure F is congruent to itself via the identity map \mbox{id}:\mathbb{R}^2\to \mathbb{R}^2, which is obviously distance-preserving. If F_1 is congruent to F_2 (via some f) then F_2 is congruent to F_1 via f^{-1}, which is distance-preserving by the prior item. Finally, if F_1 is congruent to F_2 (via some f) and F_2 is congruent to F_3 (via some g) then F_1 is congruent to F_3 via g\circ f, which is easily checked to be distance-preserving.
Problem 4

In practice the matrix for the distance-preserving linear transformation and the translation are often combined into one. Check that these two computations yield the same first two components.

a  &c  \\
b  &d
\begin{pmatrix} x  \\ y \end{pmatrix}
+\begin{pmatrix} e \\ f \end{pmatrix}
a  &c  &e \\
b  &d  &f \\
0  &0  &1
\begin{pmatrix} x  \\ y \\ 1 \end{pmatrix}

(These are homogeneous coordinates; see the Topic on Projective Geometry).


The first two components of each are ax+cy+e and bx+dy+f.

Problem 5
  1. Verify that the properties described in the second paragraph of this Topic as invariant under distance-preserving maps are indeed so.
  2. Give two more properties that are of interest in Euclidean geometry from your experience in studying that subject that are also invariant under distance-preserving maps.
  3. Give a property that is not of interest in Euclidean geometry and is not invariant under distance-preserving maps.
  1. The Pythagorean Theorem gives that three points are colinear if and only if (for some ordering of them into P_1, P_2, and P_3), \text{dist}\,(P_1,P_2)+\text{dist}\,(P_2,P_3)=\text{dist}\,(P_1,P_3). Of course, where f is distance-preserving, this holds if and only if \text{dist}\,(f(P_1),f(P_2))+\text{dist}\,(f(P_2),f(P_3))=\text{dist}\,(f(P_1),f(P_3)), which, again by Pythagoras, is true if and only if f(P_1), f(P_2), and f(P_3) are colinear. The argument for betweeness is similar (above, P_2 is between P_1 and P_3). If the figure F is a triangle then it is the union of three line segments P_1P_2, P_2P_3, and P_1P_3. The prior two paragraphs together show that the property of being a line segment is invariant. So f(F) is the union of three line segments, and so is a triangle. A circle C centered at P and of radius r is the set of all points Q such that \text{dist}\,(P,Q)=r. Applying the distance-preserving map f gives that the image f(C) is the set of all f(Q) subject to the condition that \text{dist}\,(P,Q)=r. Since \text{dist}\,(P,Q)=\text{dist}\,(f(P),f(Q)), the set f(C) is also a circle, with center f(P) and radius r.
  2. Here are two that are easy to verify: (i) the property of being a right triangle, and (ii) the property of two lines being parallel.
  3. One that was mentioned in the section is the "sense" of a figure. A triangle whose vertices read clockwise as P_1, P_2, P_3 may, under a distance-preserving map, be sent to a triangle read P_1, P_2, P_3 counterclockwise.