403/503 – Homework 5

This set is due Friday, March 26.

  1. The taxicab norm on {{\mathbb R}^2} is defined by setting {\|{\mathbf v}\|=|v_1|+|v_2|,} where {{\mathbf v}=(v_1,v_2).} Show that this is indeed a norm. As usual, we can define a distance function by setting {d({\mathbf v},{\mathbf w})=\|{\mathbf v}-{\mathbf w}\|.} Find 2 non-congruent, non-degenerate triangles with sides of length 1, 1, and 2.
  2. This exercise is related to an 80-year old open problem of Paul Erdös. Consider a unit square {H}. Inscribe in {H} exactly {n} squares with no common interior point. (The squares do not need to cover all of {H}.) Denote by {e_1,\dots,e_n} the side lengths of these squares, and define
    \displaystyle  f(n)=\max\sum_{i=1}^ne_i.

    Show that {f(n)\le\sqrt n}, and that equality holds iff {n} is a perfect square.

    (Erdös problem is to find all {n} for which {f(n)=f(n+1)}. Currently, it is only known that {f(n)<f(n+2)} for all {n}, {f(1)=f(2)=1}, {f(4)=f(5)=2}, and that if {f(n)=f(n+1)}, then {n} is a perfect square.)

  3. (This exercise comes from Jerry Shurman, Geometry of the quintic, Wiley-Interscience, 1997.)A rigid motion of {{\mathbb R}^3} is a function {R:{\mathbb R}^3\rightarrow{\mathbb R}^3} such that

    \displaystyle  \langle R({\mathbf y})-R({\mathbf x}),R({\mathbf z})-R({\mathbf x})\rangle=\langle {\mathbf y}-{\mathbf x},{\mathbf z}-{\mathbf x}\rangle

    for all vectors {{\mathbf x},{\mathbf y},{\mathbf z}\in{\mathbb R}^3.} (Here, {\langle\cdot,\cdot\rangle} is the usual inner product on {{\mathbb R}^3}.) The goal of this exercise is to show that these maps are precisely the functions of the form {R({\mathbf x})=A{\mathbf x}+{\mathbf b}} where {A} is an orthogonal matrix and {{\mathbf b}\in{\mathbb R}^3}.

    1. By definition, a {3\times 3} matrix {A} with real entries is orthogonal iff {A^TA=I,} where {A^T} is the transpose of {A}. Show that this is equivalent to stating that {A} preserves inner products, in the sense that
      \displaystyle  \langle A{\mathbf x},A{\mathbf y}\rangle=\langle{\mathbf x},{\mathbf y}\rangle

      for all {{\mathbf x},{\mathbf y}\in{\mathbb R}^3}.

    2. Show that if {R({\mathbf x})=A{\mathbf x}+{\mathbf b}} where {A} is orthogonal, then {R} is indeed a rigid motion.
    3. To prove the converse, assume now that {R:{\mathbb R}^3\rightarrow{\mathbb R}^3} is rigid.
      1. Show first that {R} is a bijection.
      2. Let {{\mathbf b}=R({\mathbf 0})} and set {S({\mathbf x})=R({\mathbf x})-{\mathbf b}.} Eventually, we want to show that {S} is a linear transformation. Begin by checking that {S({\mathbf 0})={\mathbf 0}} and that {\langle S({\mathbf x}),S({\mathbf y})\rangle=\langle {\mathbf x},{\mathbf y}\rangle} for all {{\mathbf x},{\mathbf y}\in{\mathbb R}^3}.
      3. Show that {\langle S({\mathbf x}+{\mathbf y}),S({\mathbf z})\rangle=\langle S({\mathbf x})+S({\mathbf y}),S({\mathbf z})\rangle} for all {{\mathbf x},{\mathbf y},{\mathbf z}\in{\mathbb R}^3}, and conclude that {S} preserves addition.
      4. Argue similarly that {S} preserves scalar multiplication, and is therefore linear.
      5. It follows that {S(x)=Ax} for some matrix {A}. Conclude the proof by checking that {A} is indeed orthogonal.
  4. Solve exercises 2-7, 10-14 from Chapter 6 of the textbook.

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16 Responses to 403/503 – Homework 5

  1. Amy Griffin says:

    Dr. Caicedo,
    On question #1, v_1 and v_1 are vectors right? The sum of their magnitudes gives another vector’s length… bold v. I’m just verifying that they’re not scalars…

    Also, #2: the max of a sum… does that mean that you take the largest value for the possible sums?

  2. Amy Griffin says:

    I apologize, I meant v_1 and v_2 on my first question.


  3. Hi Amy,

    1. What I meant was {\mathbf v} is a vector in {\mathbb R}^2. It therefore has two coordinates (with respect to the standard basis), and I am calling the first v_1 and the second v_2. (So, yes, v_1, v_2 are scalars.)

    2. Yes, that is what I meant, the maximum (or sup, perhaps) is taken over all possible sums.

  4. Nick Davidson says:

    Hey Dr. Caicedo,
    I think you meant to say in your last comment that {\mathbf v} \in {\mathbb R}^2.


    [*Sigh* Yes, thanks. Corrected. -A]

  5. Bailey says:

    We have not defined “non-congruent” yet. Are we talking about using the standard distance measure and using trig to show they have different angle measures?

    • Let’s be as generous as it is reasonable, and take congruent to mean same lengths, angles, areas. What I mean is, we want two “obviously different” triangles whose sides have the same lengths nonetheless. (And non-degenerate is intended to eliminate the example of a segment with a point in the middle.)

  6. Amy Griffin says:

    Dr. Caicedo,
    In the assigned problem 3 #5, To show A is orthogonal, do we prove A^T(A)=I? If so, these two matrices can get very messy with 3X3 real entries… is there another way?

    On Assigned problem 3 #1, what is required to show 1 to 1 and onto here?

    • Hi Amy,
      For 3.5, do you have another characterization of orthogonality? (Perhaps one proved earlier somewhere in problem 3?)
      I am not sure I understand your question about 3.1. Can you tell me what you mean?

  7. Amy Griffin says:

    Hi, I am not sure how to show R is a bijection. I suppose I am still unclear what a rigid motion is. I keep going through the motions of showing one thing after another, but am not sure what the big picture is for R.

    • The definition of being rigid is the equation at the beginning, \langle R(v)-R(u),R(w)-R(u)\rangle=\langle v-u,w-u\rangle for all u,v,w.

      What do you get if v=w? This says that R preserves distances. It implies (very easily) that R is 1-1 and that it is continuous.

      You may remember from vector calculus that \langle x,y\rangle=|x||y|\cos\theta, where \theta is the angle between the vectors x and y. Since R preserves distances, the equation defining rigid motions then says that the angle between R(v)-R(u) and R(w)-R(u) is the same as the angle between v-u and w-u.

      In other words: Rigid motions preserve distances and angles. That is what a rigid motion is: A continuous transformation of {\mathbb R}^3 with those two properties.

      I’ll write about surjectivity a bit later.

    • The most direct argument for surjectivity that I have thought of is as follows; there are several claims I make along the way that need to be justified, of course, but this is the skeleton:

      Let P\in{\mathbb R}^3, we want to find some P' such that R(P')=P. Pick points A,B,C in the range of R, say R(A')=A,R(B')=B,R(C')=C.

      Consider the tetrahedron ABCP. There are only two tetrahedrons in {\mathbb R}^3 that are congruent to ABCP and have A'B'C' corresponding to ABC. Say, A'B'C'P_1 and A'B'C'P_2. Then P must be the image of either P_1 or P_2.

  8. Jerry Shurman says:

    A better layout of the exercise would have been to show first that S preserves inner products and then show that S bijects. To show that S surjects one could also work in coordinates:

    Let \{e_1,e_2,e_3\} be the standard orthonormal basis in {\mathbb R}^3. Since \langle Se_i,Se_j\rangle=\langle e_i,e_j\rangle for all i,j with 1\le i\le 3 and 1\le j\le 3, then \{Se_i\mid i=1,2,3\} is an orthonormal basis.

    Next, \langle S{\mathbf x},Se_i\rangle=\langle {\mathbf x},e_i\rangle for any vector {\mathbf x} and i=1,2,3.

    So to make S hit any point, express the point as {\mathbf y}=a_1Se_1+a_2Se_2+a_3Se_3.
    Then let {\mathbf x}=a_1e_1+a_2e_2+a_3e_3.
    Then S{\mathbf x}=a_1Se_1+a_2Se_2+a_3Se_3={\mathbf y}.

    Maybe this immediately shows that S is linear, further shortening the exercise?

    • Jerry Shurman says:

      HTML ate my inner product symbols.

      [Edited: I fixed it. Thanks! -A.]

    • Hi Jerry,

      Thanks a lot!

      Very neat solution. The solution I suggested above and the one below depend a bit too much on geometric ideas that don’t seem relevant to the problem, so I like your approach much better.

    • There are a few statements that probably need more details (if this were to be turned in as homework), but this is really an effective proof.

      Thanks for sharing.

  9. Here is another way of proving surjectivity. It was suggested by my friend and colleague Ramiro de la Vega. I think that Dr. Shurman‘s solution above is the best one, but all the solutions we have exhibit different ideas which may prove useful in different contexts.

    For this approach, first prove that the image under R of a straight line is a straight line. This is the key observation.

    Surjectivity is now easy: Show now that the image under R of a plane is a plane. For this, consider three noncollinear points in a plane \Pi, and note that \Pi is the union of the lines that go through one of these three points and the opposite side in the triangle they determine.

    Finally, consider a tetrahedron, and note that any point in space is in one of the lines going through a vertex and the opposite face.

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