Question: What type of matrix normalisation is this

Could anyone tell me what type of matrix normalisation this is and it it built into Maple? I found this in a paper and Google AI said this is the process of normailsation being used.

restart

with(LinearAlgebra):

J:=Matrix([[ 1 , 0 , 0 ],
        [ 0 , 1 , 0 ],
        [ 0 , 0 , -1 ]]);

Matrix(3, 3, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = -1})

(1)

Matrix N has already been scaled, so the rows are of equal magnitude  x^2 + y^2 - z^2= k

N:=Matrix(3, 3, [[4, -3, 2], [-1/2*sqrt(70), -3/10*sqrt(70), 1/5*sqrt(70)], [1/11*sqrt(77), 6/11*sqrt(77), 2/11*sqrt(77)]])

Matrix(3, 3, {(1, 1) = 4, (1, 2) = -3, (1, 3) = 2, (2, 1) = -(1/2)*sqrt(70), (2, 2) = -(3/10)*sqrt(70), (2, 3) = (1/5)*sqrt(70), (3, 1) = (1/11)*sqrt(77), (3, 2) = (6/11)*sqrt(77), (3, 3) = (2/11)*sqrt(77)})

(2)

 

for i to 3 do
add(N[i,j]^2,j=1..2)-N[i,3]^2;
end do

21

 

21

 

21

(3)

Normalisation process to produce C

A:=N.J.N^%T

Matrix(3, 3, {(1, 1) = 21, (1, 2) = -(3/2)*sqrt(70), (1, 3) = -(18/11)*sqrt(77), (2, 1) = -(3/2)*sqrt(70), (2, 2) = 21, (2, 3) = -(27/110)*sqrt(70)*sqrt(77), (3, 1) = -(18/11)*sqrt(77), (3, 2) = -(27/110)*sqrt(70)*sqrt(77), (3, 3) = 21})

(4)

dA:=DiagonalMatrix(1/~(sqrt~(abs(Diagonal((A))))));

Matrix(3, 3, {(1, 1) = (1/21)*sqrt(21), (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = (1/21)*sqrt(21), (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = (1/21)*sqrt(21)})

(5)

C:=(dA.A.dA);

Matrix(3, 3, {(1, 1) = 1, (1, 2) = -(1/14)*sqrt(70), (1, 3) = -(6/77)*sqrt(77), (2, 1) = -(1/14)*sqrt(70), (2, 2) = 1, (2, 3) = -(9/770)*sqrt(70)*sqrt(77), (3, 1) = -(6/77)*sqrt(77), (3, 2) = -(9/770)*sqrt(70)*sqrt(77), (3, 3) = 1})

(6)

evalf(C)

Matrix(3, 3, {(1, 1) = 1., (1, 2) = -.5976143047, (1, 3) = -.6837634587, (2, 1) = -.5976143047, (2, 2) = 1., (2, 3) = -.8581163304, (3, 1) = -.6837634587, (3, 2) = -.8581163304, (3, 3) = 1.})

(7)
 

 

Download 2026-08-14_Q_What_Type_of_Matrix_Normalisation.mw

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