Hidden treasures in 2 × 2 linear systems—applications of non-orthonormal coordinate

systems.

Jon DavidsonSouthern State Community CollegeHillsboro, Ohiomath@sscc.edu

What can we learn from an ordinary linear system of two variables?

For example:

3x  −  2y  =  5−x  +  5y  =  7

The solution is x = 3, y = 2.

The coefficient matrix, A, describes a vector space in : °

2

The matrix form, , for this system is:

3 − 2− 1 5

⎡

⎣⎢

⎤

⎦⎥ 

xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥  =   5

7⎡

⎣⎢

⎤

⎦⎥

Ax =b

The right side of this equation, , is a

coordinate, , in this vector space.

57

⎡

⎣⎢

⎤

⎦⎥

32

⎡

⎣⎢

⎤

⎦⎥

The matrix form, , for this system is:

3 − 2− 1 5

⎡

⎣⎢

⎤

⎦⎥ 

xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥  =   5

7⎡

⎣⎢

⎤

⎦⎥

Ax =b

Thus we have this equation:

3  3− 1

⎡

⎣⎢

⎤

⎦⎥  +  2  − 2

5⎡

⎣⎢

⎤

⎦⎥  =   5

7⎡

⎣⎢

⎤

⎦⎥

The matrix form, , for this system is:

3 − 2− 1 5

⎡

⎣⎢

⎤

⎦⎥ 

xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥  =   5

7⎡

⎣⎢

⎤

⎦⎥

Ax =b

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 repre-sents coordinates (x, y).

For our example matrix, , the

coordinate systems represented by Au = x are as follows:

A  =   3 − 2− 1 5

⎡

⎣⎢

⎤

⎦⎥

u

v

v

x

y

u

The basis vectors for(u, v) areu = 3i – jv = – 2i + 5j.

u

v

x

yThe 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 representsinteger solutions to Au = x.

For example, coordinate (14, – 9) in (x, y) shows that (4, – 1) is the solution to this system: 3u – 2v  =  14 – u + 5v = – 9

u

v

x

y

A “unit square” in the (u, v) coordinate system is a parallel-ogram 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 and a11i  +  a21 j a12i  +  a22 j , then

area  =  absa11 a12

a21 a22

⎛

⎝⎜⎜

⎞

⎠⎟⎟ . 

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   =  

δxδu

δxδv

δyδu

δyδv

  =   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 berepresented by

transforms the unit square to the parallelogram we saw before:

S   =   0 10 0

1 01 1

⎡

⎣⎢

⎤

⎦⎥ , so

A  =   3 − 2− 1 5

⎡

⎣⎢

⎤

⎦⎥

A⋅S  =   0 30 − 1

1 − 24 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,

easier, provided you know the right substitution:

x  =   3u – 2vy = – u + 5v

u2   +  v2   =  1

This substitution turns the original equation into the unit circle:

2x2   +  2xy  +  y2   =  13 , 

Since the Jacobian, the scaling factor, is 13, the area of the ellipse is 13π.

The matrix form for the conic section is

which gives interesting avenues for exploration in its own right.

Ax2   +  Bxy  +  Cy2   =  D

x y⎡⎣

⎤⎦ 

A    B2

B2

   C

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

 xy

⎡

⎣⎢⎢

⎤

⎦⎥⎥  =  D, 

Unfortunately, I haven’t figured out a simple way to find a convenient substitution, for the general ellipse,

that would turn the equation into

in order to easily determine the area of the ellipse.

Ax2   +  Bxy  +  Cy2   =  D, 

x  =  Tu, 

u2   +  v2   =  D

det T( )  =  A     

B2

B2

    C

But if you could find the suitable x = Tu, it can be shown that

Since this is the Jacobian of the coordinate transformation, x = Tu, then it can be determined that the area of the ellipse,

area  =  Dπ

det T( )

The proof is tedious.

Another observation . . .

u

v

x

y

What is the density of the integer nodes coinciding with both(u, v) and(x, y) coordinate systems?

For this example, the density is1

13 .

This coordinate system is generated by the vector space with this ordered basis:

1− 1

⎡

⎣⎢

⎤

⎦⎥, 

24

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎩⎪

⎫⎬⎭⎪

Thus it comes from this linear system: u + 2v = x– u + 4v = y

Note that

1   2− 1   4

  =  6. 

1

6 . 

And so the density of the integer nodes is

More intuitive is observing that the area of the “unit” parallelograms is , so this rescales the number of integer nodes by that factor.det A( )

For the general system, Au = x, provided A is nonsingular and all (u, v) and (x, y) coordinates are integers, the density of integer solutions in at integer nodes is: °

2

An algebraic proof seems tricky for first time linear algebra students.

1

det A( )

This explains why integer solutions to most systems with a determi-nant of 1 or – 1 are “large” in comparison to the coefficients.

For example, consider this system:3x + 2y = 44x + 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:

34

⎡

⎣⎢

⎤

⎦⎥, 

23

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎩⎪

⎫⎬⎭⎪

The angle between 3i + 4j and 2i + 3j is about 3.2˚.

Because the determinant of the coefficient matrix for3u + 2v = x4u + 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.

For example, how does affect the unit circle?3 − 2

− 1 5⎡

⎣⎢

⎤

⎦⎥

If we let be the unit circle in (u, v), then is used to provide a substitution to turn it into (x, y) coordinates:

u2   +  v2   =  1 u  =  A− 1 x

u   =  113

  5x  +  2y( )

v  =  113

  x  +  3y( )

This gives the ellipse 2x2   +  2xy  +  y2   =  13 . 

Animating this transformation from circle to ellipse provides a little razzle-dazzle.

QuickTime™ and aGIF decompressor

are needed to see this picture.

Recall this system:3u + 4v = x2u + 3v = y

Here is its transformation of the unit circle:

QuickTime™ and aGIF 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:

for 0 ≤ t ≤ 1.

I   +  t A  −  I( )

It is interesting to show that the eigenvectors of are the same as the eigenvectors of A when t ≠ 0.

I   +  t A  −  I( )

Here are two more transformations of familiar functions. I find that students are fascinated by such transformations.

transforms this to:3 − 2

− 1 5⎡

⎣⎢

⎤

⎦⎥

1

13  x  +  3y( )  =  

1133  125x3  +  150x2y  +  60xy2   +  8y3( )

Here is the basic cubic, v  =  u3.

Here’s a transformation on v = sin u.

− 1

14  − 5x  +  y( )  =  sin − 

1

14x  −  3y( )

⎛⎝⎜

⎞⎠⎟

 .  The graph is

If you’d like a copy of this PowerPoint, pleasewrite to me (Jon Davidson) at this address:

math@sscc.edu