Download
1 / 50
500 likes | 604 Views
Polynomial Embeddings for Quadratic Algebraic Equations. Radu Balan University of Maryland, College Park, MD 20742 Math-CS Joint Lecture, Drexel University Monday April 23, 2012. Overview. Introduction: Motivation and statement of problem Invertibility Results in the Real Case
E N D
Polynomial Embeddings for Quadratic Algebraic Equations Radu Balan University of Maryland, College Park, MD 20742 Math-CS Joint Lecture, Drexel University Monday April 23, 2012
Overview • Introduction: Motivation and statement of problem • Invertibility Results in the Real Case • The Algebraic Approach: • Quasi-Linear Embeddings • Hierarchical Embeddings • Numerical Analysis • Modified Least Square Estimator • Theoretical Bounds: CRLB • Performance Analysis
1. Introduction: Motivation Inversion of Nonlinear Transformations y = A x ? knowns BA=Identity Question:What if |y| is known instead (that is, one looses the phase information) ? Where is important: X-Ray Crystallography, Speech Processing
1. Introduction: Statement of the Problem Reconstruction from magnitudes of frame coefficients a complete set of vectors (frame) for the n-dimensional Hilbert space H (H=Cn or H=Rn). Equivalence relation: x,yH, x~yiff there is a scalar z, |z|=1 so that x=zy (real case: x=y ; complex case: x=eiy). Let . Define Problems: Is an injective map? If it is, how to invert it efficiently?
H f1 x (x) fm f2 Rm
2. Invertibility Results: Real Case (1) • Real Case:K=R • Theorem[R.B.,Casazza, Edidin, ACHA(2006)] • is injective iff for any subset JF either J or F\J spans Rn. • Corollaries [2006] • if m 2n-1, and a generic frame set F, then is injective; • if m2n-2 then for any set F, cannot be injective; • if any n-element subset of F is linearly independent, then is injective; for m=2n-1 this is a necessary and sufficient condition.
Invertibility Results: Real Case (2) • Real Case:K=R • Theorem [R.B.(2012)] is injective iff any one of the following equivalent conditions holds true: • For any x,yRn, x≠0, y≠0, • There is a constant a>0 so that for all x, • RI
Invertibility Results: Real Case (3) complete set of vectors (frame) for H=Rn. One would expect that if is injective and m>2n-1 then there is a strict subset J{1,2,…,m} so that π is injective, where π:RmR|J| is the restriction to J index. However the next example shows this is not the case. Example. Consider n=3, m=6, and F the set of columns of the following matrix F = Note that for any subset J of 3 columns, either J or F\J is linearly dependent. Thus is injective but removing any column makes πnot injective.
3. The Algebraic Approach 3.1 Quasi-Linear Embeddings Example (a) Consider the real case: n=3 , m=6. Frame vectors: Need to solve a system of the form:
Then factor: Thus, we obtain:
Example (b) Consider the real case: n=3 , m=5. Frame vectors: Need to solve the system: Is it possible? How? X
Let’s square again: We obtained 5 linear equations with 6 unknown monomials. Idea: Let’s multiply again these equations (square and cross) New equations: : 15 equations New variables (monomials): : 15 unknowns
3.2 Hierarchical Embeddings Primary data: Level d embedding: , Identify: Then a homogeneous polynomial of total degree 2d.
How many monomials? Real case: number of monomials of degree 2d in n variables: Complex case: … Number of degree (d,d) in n variables: Define redundancy at level d: (R.B. [SampTA2009])
Fundamental question: How many equations are linearly independent? Recall: , Note: and The matrix is not canonical, and so is * for d>1. However its range is basis independent. We are going to compute this range in terms of a canonical matrix.
TheoremThe following hold true: (as a quadratic form) Rows of are linearly independent iff where is the mdxmd matrix given by Real case: Complex case:
Let denote the Gram matrix And for integer p. Theorem In either real or complex case: Hence for d=1, the number of independent quadratics is given by: TheoremFor d=2, Remark Note the k1=k2,l1=l2submatrix of is
3.3. Numerical analysis Results for the complex case: random frames n=3,m=6
n=16 m=64
n=32 m=128
n=4 m=16
n=4 m=14 Note 6 zero eigenvalues of instead of 5.
n=4 m=15 Number of zero eigenvalues = 20 =120-100, as expected.
4. Modified Least Square Error Estimator • Model: “Vanilla” Least-Square-Estimator (LSE): We modify this criterion in two ways: Replace X by a rank r positive matrix Y Regularize the criterion by adding a norm of Y
Optimization procedure Our approach: Start with a large and decrease its value over time; Replace one L in the inner quadratic term by a previous estimate Penalize large successive variations.
Algorithm (Part I) Step I: Initialization Step II: Iteration Step III: Factorization How to initialize? How to adapt? Convergence?
Initialization Set:
Convergence Consider the iterative process: Theorem Assume (t)t,(µt)t are monotonically decreasing non-negative sequences. Then (jt)t≥0 is a monotonically decreasing convergent sequence.
The LSE/MLE is a biased estimator. Modified CRLB for biased estimators: Asymptotically (for high SNR):
6. Performance Analysis Mintrace algorithm: Candes, Strohmer, Voroninski (2011) n=3 , m=9 , d = 1 (dlevel) , r = 2 and , decrease by 5% every step w/ saturation (subspace)
