The taxicab norm on is defined by setting where Show that this is indeed a norm. As usual, we can define a distance function by setting Find 2 non-congruent, non-degenerate triangles with sides of length 1, 1, and 2.
This exercise is related to an 80-year old open problem of Paul Erdös. Consider a unit square . Inscribe in exactly squares with no common interior point. (The squares do not need to cover all of .) Denote by the side lengths of these squares, and define
Show that , and that equality holds iff is a perfect square.
(Erdös problem is to find all for which . Currently, it is only known that for all , , , and that if , then is a perfect square.)
(This exercise comes from Jerry Shurman, Geometry of the quintic, Wiley-Interscience, 1997.)A rigid motion of is a function such that
for all vectors (Here, is the usual inner product on .) The goal of this exercise is to show that these maps are precisely the functions of the form where is an orthogonal matrix and .
By definition, a matrix with real entries is orthogonal iff where is the transpose of . Show that this is equivalent to stating that preserves inner products, in the sense that
for all .
Show that if where is orthogonal, then is indeed a rigid motion.
To prove the converse, assume now that is rigid.
Show first that is a bijection.
Let and set Eventually, we want to show that is a linear transformation. Begin by checking that and that for all .
Show that for all , and conclude that preserves addition.
Argue similarly that preserves scalar multiplication, and is therefore linear.
It follows that for some matrix . Conclude the proof by checking that is indeed orthogonal.
Solve exercises 2-7, 10-14 from Chapter 6 of the textbook.
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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?
1. What I meant was is a vector in It therefore has two coordinates (with respect to the standard basis), and I am calling the first and the second (So, yes, are scalars.)
2. Yes, that is what I meant, the maximum (or sup, perhaps) is taken over all possible sums.
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.)
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?
Thanks,
Amy
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?
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.
Thanks,
Amy
The definition of being rigid is the equation at the beginning, for all
What do you get if ? This says that preserves distances. It implies (very easily) that is 1-1 and that it is continuous.
You may remember from vector calculus that where is the angle between the vectors and . Since preserves distances, the equation defining rigid motions then says that the angle between and is the same as the angle between and
In other words: Rigid motions preserve distances and angles. That is what a rigid motion is: A continuous transformation of with those two properties.
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 we want to find some such that Pick points in the range of say
Consider the tetrahedron There are only two tetrahedrons in that are congruent to and have corresponding to Say, and Then must be the image of either or
A better layout of the exercise would have been to show first that preserves inner products and then show that bijects. To show that surjects one could also work in coordinates:
Let be the standard orthonormal basis in Since for all with and then is an orthonormal basis.
Next, for any vector and
So to make hit any point, express the point as
Then let
Then
Maybe this immediately shows that is linear, further shortening the exercise?
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.
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 of a straight line is a straight line. This is the key observation.
Surjectivity is now easy: Show now that the image under of a plane is a plane. For this, consider three noncollinear points in a plane and note that 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.
is defined by setting
where
Show that this is indeed a norm. As usual, we can define a distance function by setting
Find 2 non-congruent, non-degenerate triangles with sides of length 1, 1, and 2.
. Inscribe in
squares with no common interior point. (The squares do not need to cover all of
the side lengths of these squares, and define
, and that equality holds iff
. Currently, it is only known that
for all
,
, and that if
is a function
such that
(Here,
is the usual inner product on
where
is an orthogonal matrix and
.
matrix
where
is the transpose of
.
is indeed a rigid motion.
and set
Eventually, we want to show that
is a linear transformation. Begin by checking that
and that
for all
for all
, and conclude that
for some matrix
A kind of library 
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?
I apologize, I meant v_1 and v_2 on my first question.
Thanks,
Amy
Hi Amy,
1. What I meant was
is a vector in 
It therefore has two coordinates (with respect to the standard basis), and I am calling the first 
and the second 
(So, yes, 
are scalars.)
2. Yes, that is what I meant, the maximum (or sup, perhaps) is taken over all possible sums.
Hey Dr. Caicedo,
.
I think you meant to say in your last comment that
Nick
[*Sigh* Yes, thanks. Corrected. -A]
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.)
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?
Thanks,
Amy
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?
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.
Thanks,
Amy
The definition of being rigid is the equation at the beginning,
for all 
What do you get if
? This says that 
preserves distances. It implies (very easily) that 
is 1-1 and that it is continuous.
You may remember from vector calculus that
where 
is the angle between the vectors 
and 
. Since 
preserves distances, the equation defining rigid motions then says that the angle between 
and 
is the same as the angle between 
and 
In other words: Rigid motions preserve distances and angles. That is what a rigid motion is: A continuous transformation of
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
we want to find some 
such that 
Pick points 
in the range of 
say 
Consider the tetrahedron
There are only two tetrahedrons in 
that are congruent to 
and have 
corresponding to 
Say, 
and 
Then 
must be the image of either 
or 
A better layout of the exercise would have been to show first that
preserves inner products and then show that 
bijects. To show that 
surjects one could also work in coordinates:
Let
be the standard orthonormal basis in 
Since 
for all 
with 
and 
then 
is an orthonormal basis.
Next,
for any vector 
and 
So to make
hit any point, express the point as 
Then let
Then
Maybe this immediately shows that
is linear, further shortening the exercise?
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.
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
of a straight line is a straight line. This is the key observation.
Surjectivity is now easy: Show now that the image under
of a plane is a plane. For this, consider three noncollinear points in a plane 
and note that 
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.