1. Lecture 17 Factor Analysis

2. Syllabus Lecture 01Describing Inverse Problems Lecture 02Probability and Measurement Error, Part 1 Lecture 03Probability and Measurement Error, Part 2 Lecture 04The L 2 Norm and Simple Least Squares Lecture 05A Priori Information and Weighted Least Squared Lecture 06Resolution and Generalized Inverses Lecture 07Backus-Gilbert Inverse and the Trade Off of Resolution and Variance Lecture 08The Principle of Maximum Likelihood Lecture 09Inexact Theories Lecture 10Nonuniqueness and Localized Averages Lecture 11Vector Spaces and Singular Value Decomposition Lecture 12Equality and Inequality Constraints Lecture 13L 1, L ∞ Norm Problems and Linear Programming Lecture 14Nonlinear Problems: Grid and Monte Carlo Searches Lecture 15Nonlinear Problems: Newton’s Method Lecture 16Nonlinear Problems: Simulated Annealing and Bootstrap Confidence Intervals Lecture 17Factor Analysis Lecture 18Varimax Factors, Empircal Orthogonal Functions Lecture 19Backus-Gilbert Theory for Continuous Problems; Radon’s Problem Lecture 20Linear Operators and Their Adjoints Lecture 21Fréchet Derivatives Lecture 22 Exemplary Inverse Problems, incl. Filter Design Lecture 23 Exemplary Inverse Problems, incl. Earthquake Location Lecture 24 Exemplary Inverse Problems, incl. Vibrational Problems

3. Purpose of the Lecture Introduce Factor Analysis Work through an example

4. Part 1 Factor Analysis

5. source A ocean sediment source B s4s4 s2s2 s3s3 s1s1

6. sample matrix S S arranged row-wise but we’ll use a column vector s (i) for individual samples)

7. theory samples are a linear mixture of sources S = C F

8. theory samples are a linear mixture of sources S = C F samples contain “elements”

9. theory samples are a linear mixture of sources S = C F sources called “factors” factors contain “elements”

10. factor matrix F F arranged row-wise but we’ll use a column vector f (i) for individual factors

11. theory samples are a linear mixture of sources S = C F coefficients called “loadings”

12. loading matrix C

13. inverse problem given S find C and F so that S=CF

14. very non-unique given T with inverse T -1 if S=CF then S=[C T -1 ][TF] =C’F’

15. very non-unique so a priori information needed to select a solution

16. simplicity what is the minimum number of factors needed call that number p

17. does S span the full space of M elements? or just a p –dimensional subspace?

18. E3E3 E2E2 s1s1 s3s3 A B s1s1 s4s4 E1E1

19. we know how to answer this question p is the number of non-zero singular values

20. E3E3 E2E2 E1E1 s2s2 s3s3 A B s1s1 s4s4 v2v2 v1v1 v3v3

21. SVD identifies a subspace but the SVD factors f (i) = v (i) i=1, p not unique usually not the “best”

22. factor f (1) v with the largest singular value usually near the mean sample sample mean minimize eigenvector minimize

23. factor f (1) v with the largest singular value usually near the mean sample sample mean minimize eigenvector minimize about the same if samples are clustered

24. s3s3 f (1) s1s1 s4s4 f (2) s4s4 s3s3 s2s2 s2s2 s1s1 E3E3 E2E2 E1E1 E3E3 E2E2 E1E1 (A) (B)

25. [U, LAMBDA, V] = svd(S,0); lambda = diag(LAMBDA); F = V'; C = U*LAMBDA; in MatLab

26. [U, LAMBDA, V] = svd(S,0); lambda = diag(LAMBDA); F = V'; C = U*LAMBDA; “economy” calculation LAMBDA is M⨉M in MatLab

27. since samples have measurement noise probably no exactly singular values just very small ones so pick p for which S≈CF is an adequate approximation

28. Atlantic Rock Dataset 51.971.2514.2811.577.0211.672.120.07 50.211.4616.4110.397.4611.272.940.07 50.081.9315.611.627.6610.692.920.34 51.041.3516.49.697.2910.822.650.13 52.290.7415.068.978.1413.191.810.04 49.181.6913.9512.117.2612.3320.15 50.821.5914.2112.856.6111.252.160.16 49.851.5414.0712.246.9511.312.170.15 50.871.5214.3812.386.6911.282.110.17 (several thousand more rows) SiO 2 TiO 2 Al 2 O 3 FeO t MgO CaO Na 2 O K 2 O

29. Al 2 0 3 Ti0 2 Al 2 0 3 Si0 2 K20K20 Fe0 Mg0 Al 2 0 3 A)B) C)D)

30. singular values, λ i index, i

31. f (5) f (2) f (3) f (4) SiO 2 TiO 2 Al 2 O 3 FeO total MgO CaO Na 2 O K2OK2O

32. C2C2 C3C3 C4C4