Transcript x - Southern State Community College

Hidden treasures in 2 × 2 linear systems—
applications of non-orthonormal coordinate
systems.
Jon Davidson
Southern State Community
College
Hillsboro, Ohio
[email protected]
What can we learn from an ordinary linear system
of two variables?
For example:
3x  2 y  5
x  5y  7
The solution is x = 3, y = 2.
The matrix form, Ax  b , for this system is:
 3


1


 2

5 
 x 
 5 

  

y
7




The coefficient matrix, A, describes a vector space in
2
:
 3    2 
,

 


1
5
 
 
 
The matrix form, Ax  b , for this system is:
 3


1


 2

5 
 x 
 5 

  

y
7




 5 
The right side of this equation, 
,
is
a

7


 3 
coordinate, 
,
in
this
vector
space.

2


The matrix form, Ax  b , for this system is:
 3


1


 2

5 
 x 
 5 

  

y
7




Thus we have this equation:
 3 
  2 
 5 
3 

2







1
5
7






In order to generalize the situation, I find it easier
to change the variables to a more convenient system,
Au
=
x,
where u represents coordinates (u, v), and x represents coordinates (x, y).
 3
For our example matrix, A  

1

 2 
,
the

5 
coordinate systems represented by Au = x are as
follows:
v
u
v
y
The basis
vectors for
(u, v) are
u = 3i – j
v = – 2i + 5j.
x
u
v
y
x
u
The solution to
3u – 2v = 5
– u + 5v = 7,
which is (3, 2) in the
(u, v) coordinate
system, is found at
(5, 7) in the (x, y)
coordinate system.
Some observations . . .
Each node on the graph represents
integer solutions to Au = x.
v
y
x
u
For example,
coordinate (14, – 9)
in (x, y) shows that
(4, – 1) is the
solution to this
system:
3u – 2v = 14
– u + 5v = – 9
A “unit square” in the (u, v)
coordinate system is a parallelogram with area = 13 in the
(x, y) coordinate system.
The determinant of A is
3
 2
 1
5
 13.
It is a good exercise for students to show that in general, if two
adjacent sides of a parallelogram are represented by the vectors
a 11 i  a 21 j and a 12 i  a 22 j , then

area  abs 


a 11
a 12
a 21
a 22

 .


This gives an insight into the Jacobian determinant, used in evaluating
multiple integrations.
By adapting our original example this way,
3u – 2v = x
– u + 5v = y ,
it could be used as a coordinate transformation in evaluating a double
integral. The Jacobian is:
J

x
x
u
v
y
y
u
v

3
 2
 1
5
 13
In
the differential of area,
dA = dx dy, can be illustrated by
scaling it to a unit square.
2
The unit square can be
represented by
 0
S  
0

1
1
0
1
 3
A  

1

0 
,
so

1 
 2 

5 
transforms the unit square to the
parallelogram we saw before:
 0
A S  
0

3
1
 1
4
 2 

5 
Thus the scaling factor of the
Jacobian determinant can be
visualized in this example. The
differential of area in (u, v) is
dA  dudv  J dxdy  13 dxdy.
This makes finding the area of this
ellipse,
2x
2
 2 xy  y
2
 13 ,
easier, provided you know the right
substitution:
x = 3u – 2v
y = – u + 5v
This substitution turns the original
equation into the unit circle:
u
2
 v
2
 1
Since the Jacobian, the scaling factor, is
13, the area of the ellipse is 13π.
The matrix form for the conic section A x
 x


 A
y  
  B
 2

B 

2 

C


2
 B xy  C y
2
 D is
 x 

  D,
y


which gives interesting avenues for exploration in its own right.
Unfortunately, I haven’t figured out a simple way to find a
convenient substitution, x  Tu,
for the general ellipse,
Ax
that
would
2
 B xy  C y
turn
u
2
2
 D,
the
 v
2
 D
in order to easily determine the area of the ellipse.
equation
into
But if you could find the suitable x = Tu, it can be shown that
B
A
det T

2

B
C
2
Since this is the Jacobian of the coordinate transformation, x = Tu,
then it can be determined that the area of the ellipse,
Ax
2
 B xy  C y
area 
The proof is tedious.
2
 D , is
D
d et T

Another observation . . .
v
For this
What is the
example, the
density of the
density is
integer nodes
1
.
coinciding
1with
3 both
(u, v) and
(x, y)
coordinate
systems?
y
x
u
This
coordinate
system is
generated by
the vector
space with
this ordered
basis:
 1   2 

, 

   1   4  
Thus it comes
from this linear
system:
u + 2v = x
– u + 4v = y
Note that
1
2
 1
4
 6.
And so the
density of the
integer nodes
is
1
6
.
For the general system, Au = x, provided A is nonsingular and all
(u, v) and (x, y) coordinates are integers, the density of integer
2
solutions in
at integer nodes is:
1
det  A 
An algebraic proof seems tricky for first time linear algebra students.
More intuitive is observing that the area of the “unit” parallelograms
is det  A  , so this rescales the number of integer nodes by that factor.
This explains why integer solutions to most systems with a determinant of 1 or – 1 are “large” in comparison to the coefficients.
For example, consider this system:
3x + 2y = 4
4x + 3y = – 5
The solution is x = 22, y = – 31.
The more general system,
3u + 2v = x
4u + 3v = y ,
is based on this vector space:
 3   2 
,

 

  4   3  
The angle between 3i + 4j and 2i + 3j is about 3.2˚.
Because the determinant of the coefficient matrix for
3u + 2v = x
4u + 3v = y
is 1, any integer values for the right side, (x, y), will produce integer
solutions in (u, v).
So the density of the integer nodes is 1.
I didn’t attempt to draw this coordinate system, but here are the u and
v axes:
Another observation . . .
It can be interesting to see how the geometry induced by the
transformation A in Au = x affects familiar graphs.
 3
For example, how does 

1

 2 
 affect the unit circle?
5 
If we let u  v  1 be the unit circle in (u, v), then u  A
is used to provide a substitution to turn it into (x, y) coordinates:
2
2
u 
1
13
v 
1
13
This gives the ellipse 2 x
2
5 x
x
 2 y
 3y
 2 xy  y
2
 13 .
 1
x
Animating this transformation from circle to ellipse provides a little
razzle-dazzle.
QuickTime™ and a
GIF decompressor
are needed to see this picture.
Recall this system:
3u + 4v = x
2u + 3v = y
Here is its transformation of the unit circle:
QuickTime™ and a
GIF decompressor
are needed to see this picture.
If you have the software to produce animations (I used Maple 15), it
might be worth it to work out the procedure with your students as an
application of a time parameter.
I used this for the transformation matrices:
I  t A  I 
for 0 ≤ t ≤ 1.
It is interesting to show that the eigenvectors of I  t  A  I  are
the same as the eigenvectors of A when t ≠ 0.
Here are two more transformations of familiar functions. I find that
students are fascinated by such transformations.
Here is the basic
3
cubic, v  u .
 3


1

1
13
x
 2 
 transforms this to:
5 
 3y 
1
13
3
1 25 x
3
 150 x y  60 xy
2
2
 8y
3

Here’s a transformation on v = sin u.
The graph is 
1
14

5x  y
1


 sin  
 x  3 y  .
 14

If you’d like a copy of this PowerPoint, please
write to me (Jon Davidson) at this address:
[email protected]