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## ELECTROMAG THEORY

by: Hailey Halvorson

59

0

6

# ELECTROMAG THEORY PHYS 210A

Hailey Halvorson
UCSB
GPA 3.8

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
6
WORDS
KARMA
25 ?

## Popular in Physics 2

This 6 page Class Notes was uploaded by Hailey Halvorson on Thursday October 22, 2015. The Class Notes belongs to PHYS 210A at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 59 views. For similar materials see /class/227137/phys-210a-university-of-california-santa-barbara in Physics 2 at University of California Santa Barbara.

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Date Created: 10/22/15
Matrix rank Matrix products A 6 RM B E Rn AB 6 R7 rankAB rankB 7 dimNA rankBrankA7n g rankAB g minrankArankB dim39r dim39r dim5 A A B A B dimm7r dimnir dimp75 Roy Smith EOE 210a 41 Matrix rank Matrix products ATA and AAT Claim rankATA rankA rankAT Proof rankATA rankA 7 dimNAT RA rankA 0 A RATA RAT A RAAT 7 RA AT a NATA NA Roy Smith EOE 210a 42 Normal equations Solving Ax b The equation Ax bis consistent if there exists at least one solution Consistent case Clearly Ax b is consistent if and only if b E Choose a particular solution A92 b The general solution is the set 96mm 9296i 1421 xi NA For x E 95501 Ax A cxL 1492Aaci 1 Normal equations ATAz ATb Note that these have the same set of general solutions because NATA NA zL eNATA So for x E 95501 ATAz ATA92zi 141T14192AT1495i ATb Roy Smth EOE 210a 4 3 Normal equations Solving Ax b To get a unique solution we need I E RA and NA 0 In this case 92 AT1y1 AT 1 Note that ATA 6 RMquot and rankATA rankA n ifNA 0 So the inverse AT071 exists Inconsistent case 1 RA But for the normal equations ATAz ATb AW 6 R AT R ATA so the normal equations are always consistent Furthermore ifNA 0 then x AT1y1 ATb uniquely solves the normal equations How does this relate to the original equation Ax 1 Roy Smith EOE 210a 44 Least squares Least squares The best solution to Ax 1 De ne 5z 2 A95 7 b residual lf Ax b is consistent then 95501 gives 495501 0 But in general we want to minimize some measure of 5 Sum of squares 5zT5z AxibTAzib xTATAx72zTATbbTb 2 Ms 239 1 Objective minfz note that fz is quadratic Calculus approach Differentiate W1 t x and set equal to zero WW BM 3f 0 3 2W Barn Roy Smth EOE 210a 4 5 Least squares Least squares 7 MT T T T 395 MT T 8 7 aziAAxzAA8xi ZaxiAb Using the fact that 896 5 gives a x ZegATAx 7 ZegATb 8x 396239 Stacking the BfBxi into a vector gives 8 M ZATAx TZATb 395 Now equating this to zero implies that every minimizing solution satis es ATAz ATb lt Normal equations and the minimum achieved is min fa foam min xTATAx 7 ZzTATb bTb 12 7 zTATb Roy Smith EOE 210a 45 Least squares Converse All solutions of the normal equations minimize the LS errori Say 95501 is a solution of the normal equations7 ATAzSoln ATbi Now consider any other at E Rquot and de ne 61 x 7 95501 Then7 N96 51 IsolnTATA5z 96mm 252 IsolnTATb bTb 651413461 SEAT495501 xglnATlldz zglnATAzmln 7 255 AD 7 zzmlnATb 17 fz501n 631413461 so 95501 minimizes W Z 0 We can also see that SEAT161 0 implies that 61 E NA N ATl Which in this case means that 61 and x are also solutions of the the normal equations Conclusion 95 is a LS solution 42gt x solves the normal equations Caveat Emptor For theoretical use onlyl ATAz AT is much more likely to have numerical problems than Ax bi Roy Smth EOE 210a 4 7 Least squares Least squares example linear af ne data tting Objective Find a and 3 so that b a 37 approximates the data 1 t1 b1 Fl 1 Linear tting equations 1 tm f bm BV x x Wa A 1 Roy Smith EOE 210a 48 Least squares Least squares Fitting polynomials to data Polynomial form I a0 out 04th M71th The matrix formulation is now 1 t1 t W1 10 b1 1 t2 t3 4 11 7 b2 1 tm 73 t7 am bm W W W A x b In this case A is a Vandermonde matrix see 28 If n g m the number of data points is greater than or equal to the order of the polynomial and 12 t ifi7 j thenNA 0 Therefore the LS solution is unique and theoretically given by ATA 1ATb xopc Roy Smith EOE 210a 4 9 Similarity and bases Coordinate matrix representations see 216 Given two sets of basis vectors of the same dimension BX z1zn and By y1yn Where X spanBX and 3 spanBy The coordinate matrix representation of an operator T X a 3 is 13lele llTx1lBy lTx2lBy m T060le In the case of an identity operator TziBy ziBy We can de ne the operator P P I By ll 325 The matrix P eifectively changes the coordinates of a vector in the basis BX to its coordinates in the basis By Roy Smith EOE 210a 410 Similarity and bases Change of basis Note that P is non7singular and unique P 1 maps a vector in the basis By to its coordinates in the basis BX Matrix operators Say we have an operator A de ned in terms of the By basis UlBy AlBy lwlBy To express this in another basis apply the change of basis matrix to each of the vectors UlBy Pllex AlBylele and we can identify this as the operator in the basis B25 14le P71 AlByP Roy Smnh EOE 210a 411 Similarity and bases Similarity transforms Matrices A and B are similar if there exists an invertible matrix Q such that A Q IBQ The change of basis operation is a similarity transform on the operator s coordinate matrix Invariance under similarity The following properties amongst others are preserved by the similarity transform A Q lBQ 7 rankA rankB 7 EigenvaluesA eigenvaluesB 7 trace A traceB The trace of a matrix is de ned as trace A Z aii sum of the diagonal elements 2391 Roy Smnh EOE 210a 412

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