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”.

The following is a Quadric:

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 .

ellipsoid

+x2 -y2 -z2 = C : Hyperboloid .

hyperboloid

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

quadr_3


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 : )

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Polynomial Interpolation: Practice Versus Theory

I have posted several times, that it is possible to pre-compute a matrix, such that to multiply a set of input-samples by this matrix, will result in the coefficients of a polynomial, and that next, a fractional position within the center-most interval of this polynomial can be computed – as a polynomial – to arrive at a smoothing function. There is a difference between how I represented this subject, and how it would be implemented.

I assumed non-negative values for the Time parameter, from 0 to 3 inclusively, such that the interval from 1 … 2 can be smoothed. This might work well for degrees up to 3, i.e. for orders up to 4. But in order to compute the matrices accurately, even using computers, when the degree of the polynomial is anything greater than 3, it makes sense to assume x-coordinates from -1 … +2 , or from -3 … +3 . Because, we can instruct a computer to divide by 3 to the 6th power more easily than by 6 to the 6th power.

And then in general, the evaluation of the polynomial will take place over the interval 0 … +1 .

The results can easily be shifted anywhere along the x-axis, as long as we only do the interpolation closest to the center. But the computation of the inverse matrix cannot.

My Example

Also, if it was our goal to illustrate the system to a reader who is not used to Math, then the hardest fact to prove, would be that the matrix of terms has a non-zero determinant, and is therefore invertible, when some of the terms are negative, as it was before.

Continue reading Polynomial Interpolation: Practice Versus Theory