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The unit process in the iterative matrix diagonalization procedure known as t me

ID: 3184084 • Letter: T

Question

The unit process in the iterative matrix diagonalization procedure known as t method is a unitary transformation that operates on rows/columns i and symmetric matrix A to make aij = aji = 0. If this transformation (from basis functions Pi and Yi to ; and ) ) is written -4.9 2aj a jj-ai (a) Show that aijis transformed to zero iftan 2- (b) Show that aan remains unchanged if neither nor v is i or j, (e Find dy and a, and show that the trace of A is not changed by thetransformation. (d) Find a and ani (where is neither i nor jand show that the sum ofthe squares of the off-diagonal elements of A is reduced by the amount 2ai

Explanation / Answer

Solutions to problem 6.4.9                                                                                            14/11/2017

   Jacobi transformation and its properties

To make understanding easy, let me consider the case n=2 for (a) and (c)

The Jacobi transformation J is defined on a matrix A = [a ij ] as

    A‘   = B =   J T A J , where J is defined as rotation matrix and J = .

Substituting for J T A J , doing matrix multi plication and simplifying , we get

b 11 =   a 11 c 2 – 2 c s a 12 + a 22 s 2;      b22 = a11 s 2 + 2 a12 c s + a 22 c 2

b 12 =   a 11 cs + a12 ( c 2– s 2) - a22 cs ; b 21 = a 11 cs + a12 ( c 2– s 2) - a22 c s

where c = cos and s = sin .

Choose b 12    = b 21 = 0

This condition gives     a 11 cs + a12 ( c 2– s 2) - a22 cs   =0.

ie. (a 11 – a22 ) cos sin  + a12 cos 2  =0

ie. (1/2) (a 11 – a22 ) 2cos sin  + a12 cos 2 =0

ie. a12 cos 2 =   (1/2) (-a 11 + a22) sin  

On simplification, this gives sin / cos   = 2 a12 / (-a 11 + a22 )

                                                      ie. tan 2 a12 / (a 22 - a11 ), which proves (a)

The general case can be derived similarly with more computational effort.

(c)Trace A = sum of diagonal elements =   (a 11 + a22 )

    Trace B = b 11 + b 22 =   a 11 c 2 – 2 c s a 12 + a 22 s 2    + a11 s 2 + 2 a12 c s + a 22 c 2

                                        =   a11 (c 2 + s 2 ) + a 22 (c 2 + s 2 )

                                        =   a11 + a 22      since (c 2 + s 2 ) = cos 2 + sin 2 =1

                                        = Trace A, which proves (c)

Ie. trace is not changed by the transformation defined by J.

The solutions to (b) and (d) would be worked out separately and submitted.