The Elliptical Image of a Circle under
a Semi-inverse Matrix Operator
Donald R. Burleson, Ph.D.
Copyright © 2022 by Donald R. Burleson
All rights reserved
In a previous article (www.blackmesapress.com/Semi-inverses.htm) I introduced the concept of the (principal) semi-inverse of an invertible square matrix. In another article (www.blackmesapress.com/Diagonalizable.htm) I stated and proved a theorem to the effect that every diagonalizable nonsingular matrix is semi-invertible. This means that for any such matrix A, there is a matrix operator
Φ such that
In effect, semi-inversion raises the matrix A to the power .
For (e.g.) the 2x2 nonsingular matrix (one of the Pauli particle-spin matrices in quantum theory)
which has eigenvalues 1 and -1 and which is already in diagonal form, we have the relatively uncomplicated semi-inversion
in principal complex value.
This semi-inverse has geometric properties that have prompted me to call it the “run and hide” operator, since, for example, (thinking of the semi-inverse matrix as a transform on the xy-plane) when applied to the square with vertices (0,0), (0,1), (1,1) and (1,0) it squeezes the square down to a thin rectangle at the bottom of the original square.
When this run-and-hide operator is applied to a circle, the result is as follows.
THEOREM. For the matrix
the semi-inverse matrix
operating on a circle transforms it into an ellipse.
PROOF: For any point (x,y) in the xy-plane, the image under the specified matrix operator is
Let be the equation of any circle. Since the given matrix operator has the effect of replacing y with , for any point on the image curve for this transform it will be true that
which is the equation of an ellipse centered at This completes the proof.
A little reflection convinces one that with this semi-inverse matrix operator what is involved here is an infinite sequence of successive ellipses, of which the given circle is simply a special case encountered along the way.