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 ( I introduced the concept of the (principal) semi-inverse of an invertible square matrix. In another article ( 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 ole1.gif .

          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 ole7.gif be the equation of any circle. Since the given matrix operator has the effect of replacing y with ole8.gif , for any point on the image curve for this transform it will be true that






which is the equation of an ellipse centered at ole12.gif 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.