Post III focused on the shearing of a circle, which was centered in the Euclidean coordinate system [ECS] we worked with. The shear operation resulted in an ellipse with an inclination against the coordinate axes of our ECS. This was interesting regarding four points:
A circle, which is centered in a chosen ECS, exhibits a continuous rotational symmetry (isotropy). This obviously allows for a decomposition of a shear operation into a sequence of two affine operations in the chosen ECS: a scaling operation (with different factors along the coordinate axes) followed by a rotation (or the other way round). Equivalently: We could switch to another specific ECS which is already rotated by a proper angle against our originally chosen ECS and just perform a scaling operation there. The rotation angle is determined by the shear parameter λ. This seems to stand in some contrast to the shearing of figures with only discrete rotational symmetries: We saw for rectangles and cubes that an additional rotation was required to replace the shear operation by a sequence of scaling and rotation operations.
Points (x, y) of circles and ellipses are described by quadratic forms in two dimensions (with some real coefficients α, β, γ, δ):
\[
\alpha \,x^2 \, + \, \beta \, x \, y \, + \, \gamma \, y^2 \:=\: \delta
\]
Quadratic forms play a general role in the mathematical description of cone-sections. (Ellipses are the results of specific cone-sections.)
Ellipses also result from projections of multi-dimensional ellipsoids onto two-dimensional coordinate planes. Multi-dimensional ellipsoids are described by quadratic forms in an ECS covering the ℝn.
Hyper-surfaces for constant probability density values of multivariate normal vector distributions form multi-dimensional ellipsoids. Here we have a link to Machine Learning where key properties of certain objects are often ruled by Gaussian distributions.
From the first point we may expect that a shear operation applied to a multi-dimensional sphere will result in a multi-dimensional ellipsoid – and that such an operation could be replaced by scaling the original sphere (with different factors along n coordinate axes of a n-dimensional ECS) followed by a rotation (or vice versa). We will explicitly investigate this for a 3-dimensional sphere in the next post.
If our assumption were true we would get a first glimpse of the fact that a general multivariate standard distribution can be created by applying a sequence of distinct affine (i.e. linear) operations to a spherical probability distribution. This is discussed in detail in another post-series in this blog.
What is a bit confusing at the moment is that a replacement of a shear operation by simpler affine operations in general seems to require at least two rotations, but only one when we work with centered isotropic bodies. We come back to this point when we discuss the decomposition of a shear matrix by the so called SVD-procedure.
In the previous post of this series we have used the radius of the circle and the shearing parameter λ to derive analytical expressions for the coordinates of special points with extremal values on our ellipse
Points with maximal and minimal y-coordinate values.
Points with a maximal or minimal distance to the symmetry center of the ellipse. I.e. the end-points of the principal diameters of the ellipse.
From the fact that shearing does not change extremal values along the axis perpendicular to the sharing direction we could easily determine the lengths of the ellipse’s principal axes and the inclination angle of the longer axis with the x-axis of our Euclidean coordinate system [ECS].
What do we have in addition? In another mini-series on ellipses
I have meanwhile described how the geometry of an ellipse is related to its quadratic form and respective coefficients of a symmetric matrix. I call this matrix Aq. It forces the components of position vectors to fulfill an equation based on a quadratic polynomial. Furthermore Aq‘s eigenvalues and eigenvectors define the lengths of the ellipse’s principal axes and their inclination to the axes of our chosen ECS. The matrix coefficients in addition allow us to determine the coordinates of the points with extremal y-values on an ellipse. We will use these results later in the present post.
Objectives of this post: Shearing of a centered, rotated ellipse
In this post I want to show that shearing a given centered, but rotated original ellipse EO results in another ellipse ES with a different inclination angle and different sizes of the principal axes.
In addition we will derive the relations of the shearing parameter λS with the coefficients of the symmetric matrix \(\pmb{\operatorname{A}}_q^S \) that defines ES. I also provide formulas for the dependence of ES‘s geometrical properties on the shear parameter λS.
There are two basic prerequisites:
We must show that the application of a shear transformation to the variables of the quadratic form which describes an ellipse EO results in another proper quadratic form and a related matrix \(\pmb{\operatorname{A}}_q^S \).
The coefficients of the resulting quadratic form and of \(\pmb{\operatorname{A}}_q^S \) must fulfill a mathematical criterion for an ellipse.
We expect point 1 to be valid because a shear operation is just a linear operation.
To get some exercise we approach our goals by first looking at the simple case of shearing an axis-parallel ellipse before extending our considerations to general ellipses with an inclination angle against the coordinate axes of our chosen ECS.
This post requires Javascript to display formulas!
A centered, rotated ellipse can be defined by matrices which operate on position-vectors for points on the ellipse. The topic of this post series is the relation of the coefficients of such matrices to some basic geometrical properties of an ellipse. In the previous posts
we have found that we can use (at least) two matrix based approaches:
One reflects a combination of two affine operations applied to a unit cycle. This approach lead us to a non-symmetric matrix, which we called AE. Its coefficients ((a, b), (c, d)) depend on the lengths of the ellipses’ principal axes and trigonometric functions of its rotation angle.
The second approach is based on coefficients of a quadratic form which describes an ellipse as a special type of a conic section. We got a symmetric matrix, which we called Aq.
We have shown how the coefficients α, β, γ of Aq and a further coefficient δ of the quadratic form can be expressed in terms of the coefficients of AE. Furthermore, we have derived equations for the lengths σ1, σ2 of the ellipse’s principal axes and the rotation angle by which the major axis is rotated against the x-axis of the Euclidean coordinate system [ECS] we work with. We have also found equations for the components of the position vectors to those points of the ellipse with maximum y-values. A major result was that the eigenvalues and eigenvectors of Aq completely control the ellipse’s properties.
In this post we determine the components of the vectors to the end-points of the ellipse’s principal axes in terms of the coefficients of Aq. Afterward we shall test our formulas by a Python program and plots for a specific example.
Reduced matrix equation for an ellipse
Our centered, but rotated ellipse is defined by a quadratic form, i.e. by a polynomial equation with quadratic terms in the components xe and ye of position vectors to points on the ellipse:
The quadratic polynomial can be formulated as a matrix operation applied to position vectors ve = (xe, ye)T. With the the quadratic and symmetric matrix Aq
However, this formal aspect will not help much to solve the equations coming below. We want to describe the vectors to the principal axes’ end-points by mathematical expressions that depend on α, β, γ, δ – and both matrices will of course deliver the same results. But: There are still two different approaches to achieve our objective.
Method 1 to determine the vectors to the principal axes’ end points
My readers have certainly noticed that we have already gathered all required information to solve our task. In the first post of this series we have performed an eigendecomposition of our symmetric matrix Aq. We found that the two eigenvectors of Aq for respective eigenvalues λ1 and λ2 point along the principal axes of our rotated ellipse:
This is trivial regarding the algebraic operations, but results in lengthy (and boring) expressions in terms of the matrix coefficients. So, I skip to write down all the terms. (We do not need it for setting up ordered numerical programs.)
Remember that you could in addition replace (α, β, γ, δ) by coefficients (a, b, c, d) of matrix AE. See the first post of this series for the formulas. This would, however, produce even longer equation terms.
We pick the ye with the positive term in the following steps. (The way for the solution with the negative term in ye is analogous.) The square of ye is:
A detailed analysis also for the other ye-expression (see above) leads to further solutions for the coordinates (=vector component values) of points with extremal values for the radii. These are the end-points of the principal axes of the ellipse:
I leave it to the reader to expand the convenience variables into terms containing the original coefficients α, β, γ, δ.
Plots
It is easy to write a Python program, which calculates and plots the data of an ellipse and the special points with extremal values of the radii and extremal values of ye. The general steps which I followed were:
Step 0: Create 100 points a unit circle. Save the coordinates in Python lists (or Numpy arrays). Use Matplotlib’s plot(x,y)-function to plot the vectors.
Step 1: Create an axis-parallel ellipse with values for the axes ha = 2.0 and hb = 1.0 along the x- and the y-axis of the Euclidean coordinate system [ECS]. Do this by applying a diagonal scaling matrix Dσ1, σ2 (see the first post of this series).
Step 2: Rotate the ellipse bei π/3 (60 °). Do this by applying a rotation matrix Rπ/3 to the position vectors of your ellipse (with the help of Numpy). Alternatively, you can first create the matrices, perform a matrix multiplication and then apply the resulting matrix to the position vectors of your unit circle.
(The limiting lines have been calculated by the formulas given above.)
Step 3: Determine the coefficients of combined matrix AE = Rπ/3 ○ Dσ1, σ2
I got for the coefficients ( (a, b), (c, d) ) of AE :
A_ell =
[[ 1. -0.8660254 ]
[ 1.73205081 0.5 ]]
Step 3: Determine the coefficients of the matrix Aq by the formulas given in the first post of this series. I got
A_q =
[[ 3.25 -1.29903811]
[-1.29903811 1.75 ]]
For δ I got:
delta = 4.0
which is consistent with the length-values of the principal axes.
Step 4: Determine values for the eigenvalues λ1 and λ2 from the Aq-coefficients by the formulas given in the first post. Also calculate them by using Numpy’s
eigenvalues, eigenvectors = numpy.linalg.eig(A_q). Theory tells us that these values should be exactly λ1 = 4 and λ1 = 1. I got
Eigenvalues from A_q: lambda_1 = 4. :: lambda_2 = 1.
Step 5: Determine the components of the normalized eigenvectors with the help of numpy.linalg.eig(A_q). I got:
Components of normalized eigenvectors by theoretical formulas from A_q coefficients:
ev_1_n : -0.8660254037844386 : 0.5000000000000002
ev_2_n : 0.5000000000000001 : 0.8660254037844385
Eigenvectors from A_q via numpyy.linalg.eig():
ev_1_num : 0.8660254037844387 : -0.5000000000000001
ev_2_num : 0.5000000000000001 : 0.8660254037844387
The deviation between ev_1_n and ev_1_num is just due to a difference by -1. This is correct as the eigenvectors are unique only up to a minus-sign in all components.
Step 6: Calculate the sinus of the rotation angle of our ellipse from Aq– and Aq-coefficients. The theoretical value is sin(2 π/3) = sin(2 pi/3) = 0.8660254037844387. I got:
sin(2. * rotation angle) of major axis of the ellipse against the ECS x-axis from A_E coefficients:
sin_2phi-A_E = 0.8660254037844388
sin(2. * rotation angle) of major axis of the ellipse against the ECS x-axis from from eigenvectors of A_q:
sin_2phi-ev_A_q = 0.8660254037844387
sin(2. * rotation angle) of major axis of the ellipse against the ECS x-axis from A_q-coefficients:
sin_2phi-coeff-A_q = 0.8660254037844388
Perfect!
Step 7: Plot the end-points of the normalized eigenvectors of Aq:
Note that in our example case the end-point of the eigenvector along the minor axis must be located exactly on the elliptic curve as the ellipses minor axes has a length of b=1!
Step 8: Calculate the components of the vectors to data-points of the ellipse with maximal absolute ye-values from the Aq-coefficients given in the previous post. Plot these data-points (here in green color).
Step 9: Calculate the components of the vectors to data-points of the ellipse with maximal values of the radii with the help of the complex formulas presented in this post and plot these points in addition.
Conclusion
In this mini-series of posts we have performed some small mathematical exercises with respect to centered and rotated ellipses. We have calculated basic geometrical properties of such ellipses from the coefficients of matrices which define ellipses in algebraic form. Linear Algebra helped us to understand that the eigenvectors and eigenvalues of a symmetric matrix, whose coefficients stem from a quadratic equation (for a conic section), control both the orientation and the lengths of the ellipse’s axes completely.
This knowledge is useful in some Machine Learning [ML] context where elliptic data appear as projections of multivariate normal distributions. Multivariate Gaussian probability functions control properties of a lot of natural objects. Experience shows that certain types of neural networks may transform such data into multivariate normal distributions in latent spaces. An evaluation of the numerical data coming from such ML-experiments often delivers the coefficients of defining matrices for ellipses.
In my blog I now return to the study of with shearing operations applied to circles, spheres, ellipses and 3-dimensional ellipsoids. Later I will continue with the study of multivariate normal distributions in latent spaces of Autoencoders. For both of these topics the knowledge we have gathered regarding the matrices behind ellipses will help us a lot.
we have clarified some basic properties of shear transformations [SHT]. We got interested in this topic, because Autoencoders can produce latent multivariate normal vector distributions, which in turn result from linear transformations of multivariate standard normal distributions. When we want to analyze such latent vector distributions we should be aware transformations of quadratic forms. An important linear transformation is a shear operation. It combines aspects of scaling with rotations.
The objects we applied SHTs to were so far only squares and cubes. Both (discrete) rotational and plane symmetries of the squares and cubes were broken by SHTs. We also saw that this symmetry breaking could not be explained by a pure scaling operation in another rotated Euclidean Coordinate System [ECS]. But cubes do not have a continuous rotational symmetry. The distances of surface points of a cube to its symmetry center show no isotropy.
However, already in the first post when we superficially worked with Blender we got the impression that the shearing of a sphere seemed to produce a figure with both plane and discrete rotational symmetries – namely ellipsoids, wich appeared to be rotated. We still have to prove this, mathematically. With this post we move a first step in this direction: We will apply a shear operation to a 2D-body with perfect continuous rotational symmetry in all directions, namely a circle. A circle is a special example of a quadratic form (with respect to the vector component values). We center our Euclidean Coordinate System [ECS] at the center of the circle. We know already that this point remains a fix-point of our transformations. As in the previous post I use Python and Matplotlib to produce visual results. But we support our impression also by some simple math.
We first check via plotting that the shear operations move an extremal point of the circle (with respect to the y-coordinate) along a line ymax = const. (Points of other layers for other values yl = const also move along their level-lines.) We then have to find out whether the produced figure really is an ellipse. We do so by mathematically deriving its quadratic form with respect to the coordinates of the transformed points. Afterward, we derive the coordinate values of points with extremal y-values after the shear transformation.
In addition we calculate the position of the points with maximum and minimum distance from the center. I.e., we derive the coordinates of end-points of the main axes of the ellipse. This will enable us to calculate the angle, by which the ellipse is rotated against the x-axis.
The astonishing thing is that our ellipse actually can be created by a pure scaling operation in a rotated ECS. This seems to be in contrast to our insight in previous posts that a shear matrix cannot be diagonalized. But it isn’t … It is just the rotational symmetry of the circle that saves us.
Shearing a circle
We define a circle with radius r = a = 2.
I have indicated the limiting line at the extremal y-values. From the analysis in the last post we expect that a shear operation moves the extremal points along this line.
We now apply a shearing matrix with a x/y-shearing parameter λ = 2.0
Thus, we have indeed produced a rotated ellipse! We see this from the fact that the term mixing the xs and the yl coordinates does not vanish.
Position of maximum absolute y-values
We know already that the y-coordinates of the extremal points (in y-direction) are preserved. And we know that these points were located at x = 0, y = a. So, we can calculate the coordinates of the shifted point very easily:
In our case this gives us a position at (4, 2). But for getting some experience with the quadratic form let us determine it differently, namely by rewriting the above quadratic equation and by a subsequent differentiation. Quadratic supplementation gives us:
Let us also find the position of the end-points of the main axes of the ellipse. One method would be to express the ellipse in terms of the coordinates (xs, ys), calculate the squared radial distance rs of a point from the center and set the derivative with respect to xs to zero.
The “problem” with this approach is that we have to work with a lot of terms with square roots. Sometimes it is easier to just work in the original coordinates and express everything in terms of (x, y):
Plot of main axes, their end-points and of the points with maximum y-value
The coordinate data found above help us to plot the respective points and the axes of the produced ellipse. The diameters’ end-points are plotted in red, the points with extremal y-value in green:
It becomes very clear that the points with maximum y-values are not identical with the end-points of the ellipse’s main symmetry axes. We have to keep this in mind for a discussion of higher dimensional figures and vector distributions as multidimensional spheres, ellipsoids and multivariate normal distributions in later posts.
Rotated ECS to produce the ellipse?
The plot above makes it clear that we could have created the ellipse also by switching to an ECS rotated by the angle α. Followed by a simple scaling in x- and y-direction by the factors as and bs in the rotated ECS. This seems to be a contradiction to a previous statement in this post series, which said that a shear matrix cannot be diagonalized. We saw that in general we cannot find a rotated ECS, in which the shear transformation reduces to pure scaling along the coordinate axes. We assumed from linear algebra that we in general need a first rotation plus a scaling and afterward a second different rotation.
But the reader has already guessed it: For a fully rotation-symmetric, i.e. isotropic body any first rotation does not change the figure’s symmetry with respect to the new coordinate axes. In contrast e.g. to squares or rectangles any rotated coordinate system is as good as any other with respect to the effect of scaling. So, it is just scaling and rotating or vice versa. No second rotation required. We shall in a later post see that this holds in general for isotropically shaped bodies.
Conclusion
Enough for today. We have shown that a shear transformation applied to a circle always produces an ellipse. We were able to derive the vectors to the points with maximum y-values from the parameters of the original circle and of the shear matrix. We saw that due to the circle’s isotropy we could reduce the effect of shearing to a scaling plus one rotation or vice versa. In contrast to what we saw for a cube in the previous post.
