M Theory Lesson 106

Let $\omega$ be the primitive cube root of unity. The determinant of a $3 \times 3$ complex circulant


is given by

$(X + Y + Z)(X + \omega Y + \omega^{2} Z)(X + \omega^{2} Y + \omega Z)$
$= X^3 + Y^3 + Z^3 – 3XYZ$

On setting the determinant to 1 (another choice of normalisation) the inverse of the circulant takes the simple form

(XX – ZY)(ZZ – XY)(YY – XZ)
(YY – XZ)(XX – ZY)(ZZ – XY)
(ZZ – XY)(YY – XZ)(XX – ZY)

which is again a 1-circulant since matrix multiplication is closed for this set. For 2-circulants, on the other hand, inverses can be 2-circulant. An inverse of an idempotent will also be idempotent.

5 Responses so far »

  1. 1

    CarlBrannen said,

    It’s late, and I’m thinking way too much about control logic for an ethanol plant, but something on the right side of my brain says that the primitive idempotent circulant matrices “P” I’ve been playing with do not have inverses.

    For instance, 3P =
    1 1 1
    1 1 1
    1 1 1.

    Probably this comes about because I believe that they can be transformed (by S P S^-1 ) into the three primitive idempotent diagonal matrices that look like this:

    1 0 0
    0 0 0
    0 0 0

    which clearly have no inverse. I think that the only idempotent matrix with an inverse is the unit matrix. Gosh, it will be embarassing if I screwed this up.

    Meanwhile, my buddy is signing for the $30 million loan tomorrow and we are underway to build the ethanol plant. I promised to design the electrical controls (be afraid, be very afraid, a misprogrammed 36 million gallon ethanol distillation plant burns very well).

    And I decided to do the NNLO calc for the Mass snuark term using the traditional technique of vacuum and creation and annihilation operators. That means that I have to first write a blog post that describes how to convert from snuark (geometric) QFT to standard QFT. Schwinger showed how to do this in the 1950s. The standard QFT is easier to program.

  2. 2

    Kea said,

    Good luck with the plant! The general idempotency condition (for ordinary matrix multiplication here) gives

    X = XX + 2YZ
    Y = ZZ + 2XY
    Z = YY + 2XZ

    and so the inverse circulant has first term

    XXXX + 2YYZZ – 2XYYY – 2XZZZ

    but using the det=1 rule this becomes … whatever … still doesn’t look like 1 or 0 to me, but then these aren’t primitive idempotents. Besides, I’m far more likely to have made a stupid mistake than you have.

  3. 3

    Kea said,

    The det=1 cubic in X=X(Y,Z) at Y=Z=0 (the identity solution) has discriminant 27, which fits with the three solutions X=1,omega,omega^2 using cube roots of unity. The general discriminant is

    D= 27[YYYYYY + ZZZZZZ + 5YYYZZZ – YYY – ZZZ + 1]

    which will be positive for positive real Y and Z sufficiently large, and hence there are three distinct solutions for X for generic (Y,Z).

    However, this doesn’t cover the case when Y and Z are both complex.

  4. 4

    Kea said,

    Arrgh. Sorry! I was looking at circulants without the idempotency condition ….

  5. 5

    CarlBrannen said,

    Our engineer responsible for the Programmable Control Logic at our plant officially quit today and now I am responsible for it. Fortunately, it appears we’ll have an experienced guy design it, but he’s not working on it yet and I have to make decisions on the system, etc.

    I bought a 1000 page book on the subject at the university bookstore yesterday. It’s actually very simple theory. They’re programmed in a bizarre language called Ladder Logic that is a throw-back to the days of relays.

    What’s worrying me is that the whole book doesn’t mention anything about simulation. My buddy and I are wondering if this is an oversight, or if they build plants and turn them on without first simulating their “PCL”. Uh, electronics engineers tend to look down their noses a bit at chemical engineers; I hope we’re being too harsh and there are standard tools for simulation.

    I’ve dreaded the day we turn this thing on ever since we bought it. I guess nothing really awful can happen until the ethanol output tanks start filling up.

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