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Instabilities of SVD in Small Eigenvalues

This article discusses the instabilities of Singular Value Decomposition (SVD) in small eigenvalues, which can be sensitive to small amounts of noise and may be indistinguishable from 0. It explores methods such as Truncated SVD (TSVD) to remove small eigenvalues in order to stabilize the solution.

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Instabilities of SVD in Small Eigenvalues

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  1. Instabilities of SVD Small eigenvalues -> m+ sensitive to small amounts of noise Small eigenvalues maybe indistinguishable from 0 Possible to remove small eigenvalues to stabilize solution -> Truncated SVD, TSVD Condition number cond(G)=s1/sk

  2. TSVD Example: removing instrument response g0 t exp(-t/T0) (t≥0) g(t)= 0 (t<0) v(t)=∫g(t-)mtrue()d (recorded acceleration) Problem: deconvolving g(t) from v(t) to get mtrue (ti-tj)exp[-(ti-tj)]/T0t (tj≥ti) d=Gm, Gi,j= 0 (tj<ti)  -

  3. TSVD time [-5,100]s t=0.5s -> G with m=n=210 Singular values 25.3->0.017, Cond(G)~1480 e.g., 1/1000 noise creates instability.. True signal: mtrue(t)=exp[-(t-8)2/22] + 0.5 exp[-(t-25)2/22 ]

  4. TSVD dtrue=Gmtrue m=VS-1UTdtrue

  5. TSVD dtrue=Gmtrue m=VS-1 UT(dtrue+), =N(0,(0.05 V)2) Solution fits data perfectly, but worthless…

  6. TSVD dtrue=Gmtrue m=VpSp-1 UpT(dtrue+), =N(0,(0.05 V)2) Solution for p=26 (removed 184 eigenvalues!)

  7. Nonlinear Regression Linear regression: Now we know (e.g., LS) Assume a nonlinear system of m eq and m unknowns F(x)=0 … What are we going to do?? We will try to find a sequence of vectors x0, x1, … that will converge toward a solution x* Linearize: (assume that F is continuously differentiable)

  8. Nonlinear Regression Linear regression: Now we know (e.g., LS) Assume a nonlinear system of m eq and m unknowns F(x)=0 … What are we going to do?? We will try to find a sequence of vectors x0, x1, … that will converge toward a solution x* Linearize: (assume that F is continuously differentiable) F(x0+x)≈F(x0)+F(x0)x where F(x0) is the Jacobian

  9. Nonlinear Regression Assume that the x puts us at the unknown solution x*: F(x0+x)≈F(x0)+F(x0)x=F(x*)=0 -F(x0) ≈ F(x0)x = Newton’s Method! F(x)=0, initial solution x0. Generate a sequence of solutions x1, x2, …and stop if the sequence converges to a solution with F(x)=0. Solve -F(xk) ≈ F(xk)x (fx, using Gaussian elimination). Let xk+1=xk+x. let k=k+1

  10. Properties of Newton’s Method If x0 is close enough to x*, F(x) is continuously differentiable in a neighborhood of x*, and F(x*) is nonsingular, Newton’s method will converge to x*. The convergence rate is ||xk+1-x*||2≤c||xk-x*||22

  11. Newton’s Method applied to a scalar function Problem: Minimize f(x) If f(x) is twice continuously differentiable f(x0+x)≈f(x0)+f(x0)Tx+1/2 xT2f(x0) x where f(x0) is the gradient and 2f(x0) is the Hessian

  12. Newton’s Method applied to a scalar function A necessary condition for x* to be a minimum of f(x) is that f(x*)=0. In the vicinity of x0 we can approximate the gradient as f(x0+x)≈f(x0)+2f(x0)Tx (eq 9.8 - higher-order terms) Setting the gradient to zero (assuming x0+x puts us at x*) we get -f(x0) ≈ 2f(x0)Tx, which is Newton’s method for minimizing f(x): Twice differentiable function f(x), initial solution x0. Generate a sequence of solutions x1, x2, …and stop if the sequence converges to a solution with f(x)=0. Solve -f(xk) ≈ 2f(xk)x Let xk+1=xk+x. 3. let k=k+1

  13. Newton’s Method applied to a scalar function Is the same as solving a nonlinear system of equations applied to f(x)=0, so therefore If x0 is close enough to x*, f(x) is twice continuously differentiable in a neighborhood of x*, and there is a constant  such that ||2f(x)-2f(y)||2≤||x-y||2 for every y in the neighborhood, and 2f(x*) is positive definite, and x0 is close enough to x*, then Newton’s method will converge quadratically to x*

  14. Newton’s Method applied to LS Not directly applicable to most nonlinear regression and inverse problems (not equal # of model parameters and data points, no exact solution to G(m)=d). Instead we will use N.M. to minimize a nonlinear LS problem, e.g. fit a vector of n parameters to a data vector d. f(m)=∑ [(G(m)i-di)/i]2 Let fi(m)=(G(m)i-di)/I i=1,2,…,m, F(m)=[f1(m) … fm(m)]T So that f(m)= ∑ fi(m)2 f(m)=∑ fi(m)2] m i=1 m i=1 m i=1

  15. Newton’s Method applied to LS f(m)j=∑ 2fi(m)jF(m)j] f(m)=2J(m)TF(m), where J(m) is the Jacobian f(m)j=∑ fi(m)2] = ∑ Hi(m), where Hi(m) is the Hessian of fi(m)2 Hij,k(m)= m i=i m i=i m i=i

  16. Newton’s Method applied to LS f(m)=2J(m)TJ(m)+Q(m), where Q(m)=∑ fi(m) fi(m) Gauss-Newton (GN) method ignores Q(m), f(m)≈2J(m)TJ(m), assuming fi(m) will be reasonably small as we approach m*. That is, NM: Solve -f(xk) ≈ 2f(xk)x f(m)j=∑ 2fi(m)jF(m)j, i.e. J(mk)TJ(mk)m=-J(mk)TF(mk) m i=i

  17. Newton’s Method applied to LS Levenberg-Marquardt (LM) method uses [J(mk)TJ(mk)+I]m=-J(mk)TF(mk) ->0 : GN ->large, steepest descent (SD) (down-gradient most rapidly). SD provides slow but certain convergence. Which value of  to use? Small values when GN is working well, switch to larger values in problem areas. Start with small value of , then adjust.

  18. Statistics of iterative methods Cov(Ad)=A Cov(d) AT (d has multivariate N.D.) Cov(mL2)=(GTG)-1GT Cov(d) G(GTG)-1 Cov(d)=2I: Cov(mL2)=2(GTG)-1 However, we don’t have a linear relationship between data and estimated model parameters for the nonlinear regression, so cannot use these formulas. Instead: F(m*+m)≈F(m*)+J(m*)m Cov(m*)≈(J(m*)TJ(m*))-1 ri=G(m*)i-di s=[∑ ri2/(m-n)] Cov(m*)=s2(J(m*)TJ(m*))-1

  19. Implementation Issues Explicit (analytical) expressions for derivatives Finite difference approximation for derivatives When to stop iterating?

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