## An oversight which I made, in an earlier posting: Matrices with Negative Determinants.

One of the subjects which I have written about a number of times, especially in This Posting, is the use of ‘rotation matrices’, and what their determinant is. This subject actually requires some understanding of Linear Algebra to be understood in turn. But it also requires just a bit more insight, into what the equations stand for.

A matrix can exist, the columns of which are mutually perpendicular – i.e., orthogonal – in addition to being unit vectors each. What I wrote was that, in such a case, the determinant of the matrix would equal (+1), and that its transpose can be used, in place of computing its inverse.

Such a matrix can be used to rotate objects that are distinctly not rectangular in appearance, but rotate them nonetheless, in computer games, CGI, etc.

A situation which I had overlooked was, that the determinant of such a matrix could also be (-1). And if it is, then to apply this matrix to a 3D system of coordinates has as effect:

• To convert between a right-handed coordinate system and a left-handed coordinate system accurately, or
• To derive a model that is the mirror-image of the original model.

What tends to happen in Scientific Computing, as well as in certain other areas, is that right-handed coordinate systems are often used, and left-handed coordinates less-frequently so. Yet, left-handed coordinate systems are still used. And so, if that is the case, this conversion will need to take place eventually, and no longer counts as a rotation. I.e., it has been observed that, if a right-handed helix is rotated whichever way, it stays a right-handed helix. Well, if such an orthonormal matrix with a determinant of (-1) is applied to its model coordinates, then it will become a left-handed helix…

Dirk

## About wanting to use the Transpose of a matrix, instead of the Inverse.

There is a subject which I’ve mentioned in scant ways, in other postings, but which I have not yet written a posting about, directly.

A situation can arise in which we have a rotation matrix, and where we’d like to invert the rotation it defines, but where the task of actually computing the inverse of the matrix would be too expensive computationally. And so the question can come up, of whether to use the transpose will also work.

The answer is, that in one specific situation the transpose can be used. This is, if each column of the matrix is a unit vector, and if all the columns are mutually orthogonal. If that’s the case, then the determinant of the matrix will also be equal to (1), and the transpose will be equal to the inverse.

Simultaneously, this will then also be true for all the rows of (either) matrix.

Therefore, when I wrote that a Perpendicular matrix could be computed, in case the original matrix defined a quadric, and that the transpose of that perpendicular matrix can be used, as if that was its inverse, I was describing a special situation, in which the perpendicular matrix would have been orthonormal to begin with.

In case a matrix which defines a conversion of coordinates is not orthonormal, then this approach will not work, and its inverse must be used. And I think that the resulting matrix is also called ‘The Jordan Product’ of an original matrix. My exercise did not need this.

Dirk

## The Difference Between a Quartic, and a Quadric

I’ve talked to people who did not distinguish, between a Quartic, and a Quadric.

The following is a Quartic:

y = ax4 + bx3 + cx2 + dx + e

It follows in the sequence from a linear equation, through a quadratic, through a cubic, to arrive at the quartic. What follows it is called a “Quintic”.

a1 x2 + a2 y2 + a3 z2 +

a4 (xy) + a5 (yz) + a6 (az) +

a7 x + a8 y + a9 z – C = 0

The main reason quadrics are important, is the fact that they represent 3D shapes such as Hyperboloids, Ellipsoids, and Mathematically significant, but mundanely insignificant shapes, that radiate away from 1 axis out of 3, but that are symmetrical along the other 2 axes.

If the first-order terms of a quadric are zero, then the mixed terms merely represent rotations of these shapes, while, if the mixed terms are also zero, then these shapes are aligned with the 3 axes. Thus, if (C) was simply equal to (5), and if the signs of the 3 single, squared terms, by themselves, are:

+x2 +y2 +z2 = C : Ellipsoid .

+x2 -y2 -z2 = C : Hyperboloid .

+x2 +y2 – z2 = C : ‘That strange shape’ .

The way in which quadrics can be manipulated with Linear Algebra is of some curiosity, in that we can have a regular column vector (X), which represents a coordinate system, and we can state the transpose of the same vector, (XT), which forms the corresponding row-vector, for the same coordinate system. And in that case, the quadric can also be stated by the matrix product:

XT M X = C

(Updated 1/13/2019, 21h35 : )