What Does Subtracting the Identity from a Transformation Mean in Nomenclature?

  • Thread starter Thread starter pamparana
  • Start date Start date
  • Tags Tags
    Nomenclature
pamparana
Messages
123
Reaction score
0
Hello everyone,

I have a question about nomenclature and it may be a bit more than simple nomenclature as I am not quite sure I understand it:

So, I am reading a paper and going through it very slowly as it is quite intense. But one of the line is as follows:

Suppose the transformation s maps a point p to the point s(p). Let u = s - I be the displacement field associated with the transformation s.

I am confused as to what s - I represents. The subtracting of the identity from a transformation seems strange to me.

I always thought the displacement field can be simply represented for each particle by associating a vector with each particle...

I hope someone can clarify this doubt for me.

Many thanks,

Luc
 
Mathematics news on Phys.org
I guess, they try to describe the vector by saying the vector is defined by its starting point and end point u = (x_2-x_1, y_2-y_1) and s((x_1,y_1)) = (x_2,y_2).

They probably used the notation (s-I)(p)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top