Download presentation

Presentation is loading. Please wait.

1
Lecture 20 SVD and Its Applications Shang-Hua Teng

2
Every symmetric matrix A can be written as Spectral Theorem and Spectral Decomposition where x 1 …x n are the n orthonormal eigenvectors of A, they are the principal axis of A. x i x i T is the projection matrix on to x i !!!!!

3
Singular Value Decomposition Any m by n matrix A may be factored such that A = U V T U: m by m, orthogonal, columns V: n by n, orthogonal, columns : m by n, diagonal, r singular values

4
The Singular Value Decomposition r = the rank of A = number of linearly independent columns/rows AU VTVT m x n m x m m x n n x n · · = 0 0

5
SVD Properties U, V give us orthonormal bases for the subspaces of A: –1st r columns of U: Column space of A –Last m - r columns of U: Left nullspace of A –1st r columns of V: Row space of A –1st n - r columns of V: Nullspace of A IMPLICATION: Rank(A) = r

6
The Singular Value Decomposition · · AU VTVT = 0 0 A U VTVT m x n m x r r x r r x n = 0 0 m x n m x m m x n n x n

7
Singular Value Decomposition where u 1 …u r are the r orthonormal vectors that are basis of C(A) and v 1 …v r are the r orthonormal vectors that are basis of C(A T )

8
SVD Proof Any m x n matrix A has two symmetric covariant matrices (m x m) AA T (n x n) A T A

9
Spectral Decomposition of Covariant Matrices (m x m) AA T =U U T –U is call the left singular vectors of A (n x n) A T A = V V T –V is call the right singular vectors of A Claim: are the same

10
Singular Value Decomposition Proof

11
All Singular Values are non Negative

12
Row and Column Space Projection Suppose A is an m by n matrix that has rank r and r << n, and r << m. –Then A has r non-zero singular values –Let A = U V T be the SVD of A where S is an r by r diagonal matrix –Examine:

13
The Singular Value Projection · A U VTVT m x n m x r r x r r x n = 0 0

14
Therefore Rows of U are r dimensional projections of rows of A Columns of V T are r dimensional projections of columns of A So we can compute their distances or dot products in a lower dimensional space

15
Eigenvalues and Determinants Product law: Summation Law: Both can be proved by examining the characteristic polynomial

16
Eigenvalues and Pivots If A is symmetric the number of positive (negative) eigenvalues equals to the number of positive (negative) pivots A = LDL T Topological Proof: scale down the off-diagonal entries of L continuously to 0, i.e., moving L continuously to I. Any change sign in eigenvalue must cross 0

17
Next Lecture Dimensional reduction for Latent Semantic Analysis Eigenvalue Problems in Web Analysis

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google