How to define an ellipse by the eigendecomposition of its transformation matrix?

I'm trying to establish the correct setup for defining an ellipse as a 'stretch' of a circle and a rotation of the result. For simplicity assume the centres of the circle and therefore the ellipse to be so that the equation of the ellipse is

Now let the eigenvalues of be and so that the 'stretch' matrix is

and let the rotation, counter-clockwise, of an angle θ from the x-axis be achieved by the transformation

.

Since , is an admissible matrix of eigenvectors and it should be possible to express as the product of its eigendecomposition by

.

However, when I perform the reverse operation in practice, clearly something in the above is not correct, but I'm not sure what it is. The code snippet below illustrates the issue for and , . What is it I'm getting wrong?

>> theta = pi/4
theta =
       0.7854
>> R = [cos(theta) sin(theta); -sin(theta) cos(theta)]
R =
      0.70711      0.70711
     -0.70711      0.70711
>> S = [1 0; 0 4]
S =
     1     0
     0     4
>> A = R*S*R'
A =
          2.5          1.5
          1.5          2.5
>> [V,D] = eig(A)
V =
     -0.70711      0.70711
      0.70711      0.70711
D =
     1     0
     0     4
>> V*D*V'-A
ans =
  -4.4409e-16  -4.4409e-16
  -4.4409e-16  -4.4409e-16
>> 

[Torsten]

And why do you think it's not correct ? Because you get an error in the order of 1e-16 ? That's numerical garbage.

[RickyBoy]

Of course that's a precision issue. It's the signs in the eigenvector that I question.

[RickyBoy]

I simply put the comparision at the end to demonstrate that

with therefore .

[John D'Errico]

It appears you are either questioning why the result is not exactly zero in double precision arithmetic, or you are questioning the sign of an eigenvector.

So why are you questioning the sign of an eigenvector? Do you recognize that you can multiply an eigenvector by any constant, and it is still an eigenvector? -1 is a constant, which does not change the norm of the eigenvector. So the eigenvectors of a matrix are not uniquely determined.

[Torsten]

Yes, even normalized eigenvectors to a one-dimensional eigenspace are only unique up to a factor of +/- 1...

And the usual form of a rotation matrix is [cos(theta) -sin(theta);sin(theta) cos(theta)].

theta = pi/4

theta = 0.7854

R = [cos(theta) -sin(theta); sin(theta) cos(theta)]

R = 2×2

0.7071 -0.7071 0.7071 0.7071

S = [1 0; 0 4]

S = 2×2

1 0 0 4

A = R*S*R'

A = 2×2

2.5000 -1.5000 -1.5000 2.5000

[V,D] = eig(A)

V = 2×2

-0.7071 -0.7071 -0.7071 0.7071

D = 2×2

1 0 0 4

V*D*V'-A

ans = 2×2

1.0e-15 * -0.4441 0.4441 0.4441 -0.4441

[RickyBoy]

OK, thank you both, your comments taken on board. I wonder if you might be kind enough to demonstrate how you would plot the ellipse in Matlab using the elements of and as given above? My attempts at the very least result in a rotation in the opposite direction expected!! Geometry was never my strongest suit! Sadly.

[William Rose]

@RickyBoy, that is a very nice compact and elegant bit of code from @Torsten.

In case you are still wondering about the sign in the rotation matrix, in which you had

R1=[cos(t), sin(t); -sin(t), cos(t)]

and @Torsten had

R2=[cos(t), -sin(t); sin(t), cos(t)]

The difference is that

y1=R1*x is equivalent to rotating the axes CCW by angle t. y1 is x, expressed in terms of the rotated axes.

y2=R2*x is equivalent to rotating the points CCW by angle t. y2 is the rotated x. The axes are unaltered.

By the way, when viewing plots of ellipses, you may want to use axis equal so that the aspect ratio is correctly represented.

[RickyBoy]

Thank you, appreciate your comments.

Answer 1

[Torsten]

phi = linspace(0,2*pi,100);
S = [1 0;0 4];
xy = S*[cos(phi);sin(phi)];
theta = pi/4;
Sxy = [cos(theta) -sin(theta);sin(theta) cos(theta)]*xy;
hold on
plot(xy(1,:),xy(2,:))
plot(Sxy(1,:),Sxy(2,:))
hold off

[RickyBoy]

Thank you, most helpful! I can't mark this as an accepted answer because it was submitted as a comment but if you submit it as an answer I will mark it as the accepted one. In any case, I appreciate your engagement.