User Name Remember Me? Password

 Linear Algebra Linear Algebra Math Forum

 October 31st, 2018, 06:54 PM #1 Member   Joined: Feb 2018 From: Canada Posts: 46 Thanks: 2 Polar decomposition I am having trouble with the fist part of this question. The polar decomposition of a real invertible matrix $A\in R^{n*n}$ is: $A=QP$ where $Q \in R^{n*n}$ is orthogonal and $P\in R^{n*n}$ symmetric positive-definite. Prove that it always exist. What is $cond_2(A)$ in terms of its polar factors $(Q \text{ and } P)$? For the second part, $cond_2(A) = \|A\|_2\|A^{-1}\|_2 = \|QP\|_2\|P^{-1}Q^{-1}\|_2 \leq \|Q\|_2\|P\|_2\|P^{-1}\|_2\|Q^{-1}\|_2 = \|P\|_2\|P^{-1}\|_2 \text{ since Q is orthogonal}$ Can someone hep me with the first part. I think we can start with singular value decomposition as $A = U\Sigma V^{T}$ then transform this into the polar decomposition. November 30th, 2018, 05:29 PM #2 Member   Joined: Feb 2018 From: Canada Posts: 46 Thanks: 2 The (real) SVD of $A = U\Sigma V^{T}$ always exists. Inserting $I=V^{T}V$ in between $U$ and $\Sigma$, we find: $A = \underbrace{UV^{T}}_{=: Q}\underbrace{V\Sigma V^{T}}_{=:P},$ Since A is invertible, $A^{-1} = V\Sigma ^{-1}U^{T}$, which implies that $\Sigma$ is invertible as well. Since the singular values of A are positive, the matrix $V\Sigma V^{T}$ is symmetric positive-definite as it is a spectral decomposition with orthogonal eigenvectors and positive eigenvalues, and the product $UV^{T}$ is clearly orthogonal. We must find a symmetric $P$ and orthogonal $Q$. Thus: $A^{T}A = P^{T}Q^{T}QP = P^{T}P = P^2.$ Since A is invertible, $A^{T}A$ is symmetric positive-definite and it has a spectral decomposition $V\Lambda V^{T}$ with orthogonal eigenvectors. Thus $P = \sqrt{A^{T}A} = \sqrt{V\Lambda V^{T}} = V\Lambda^{\frac{1}{2}} V^{T},$ is also symmetric positive-definite. Finally, $Q = V\Lambda^{-\frac{1}{2}}V^{T}$ is orthogonal, since: \begin{aligned} Q^{T}Q &= V\Lambda^{-\frac{1}{2}}V^{T}A^{T}AV\Lambda^{-\frac{1}{2}}V^{T} = V\Lambda^{-\frac{1}{2}}V^{T}V\Lambda V^{T}V\Lambda^{-\frac{1}{2}}V^{T}, \\ &= V\Lambda^{-\frac{1}{2}}\Lambda \Lambda^{-\frac{1}{2}}V^{T} = VV^{T} = I. \end{aligned} Since Q is orthogonal: \begin{aligned} cond_2(A) &= \|A\|_2 \|A^{-1}\|_2 \\ &= \|QP\|_2 \|(QP)^{-1}\|_2 \\ &= \|QP\|_2 \|P^{-1}Q^{-1}\|_2 \\ &\leq\|Q\|_2\|P\|_2\|P^{-1}\|_2\|Q^{-1}\|_2 \\&= \|P\|_2\|Q\|_2\|Q^{-1}\|_2 |P^{-1}\|_2 \\ &= \|P\|_2\|P^{-1}\|_2 \\ &= cond_2(P) \end {aligned} On the other hand, $A = QP \rightarrow Q^{T}A = Q^{T}QP \rightarrow Q^{T}A = P$ \begin{aligned} \text{then} \quad cond_2(P) &= \|P\|_2 \|P^{-1}\|_2 \\ &= \|Q^{T}A\|_2 \|(Q^{T}A)^{-1}\|_2 \\ &= \|Q^{T}A\|_2 \|A^{-1}Q\|_2 \\ &\leq\|Q^{T}\|_2\|A\|_2\|A^{-1}\|_2\|Q\|_2 \\&= \|A\|_2\|Q\|_2\|Q^{-1}\|_2 |A^{-1}\|_2 \\ &= \|A\|_2\|A^{-1}\|_2 \\ & = cond_2(A) \end{aligned} Thus, $cond_2(A) = cond_2(P)$. Last edited by Shanonhaliwell; November 30th, 2018 at 05:31 PM. December 2nd, 2018, 07:36 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics Your 1st post seems fine though its a bit of overkill. Note that if $A,B$ are linear operators, then $cond(AB) = cond(A)cond(B)$ which is trivial to prove. Then, since $cond(U) = 1$ for any unitary matrix, it follows that when $A = UP$, then $cond(A) = cond(P)$. In your 2nd post you seem to be done in the first expression so it isn't clear what you are trying to do. I am also mildly suspicious that your teacher has in mind you begin with an SVD since this is typically viewed as a generalized form of the polar decomposition. Thanks from Shanonhaliwell Tags decomposition, polar Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post life24 Linear Algebra 1 July 8th, 2016 12:41 PM mick7 Linear Algebra 3 December 25th, 2013 10:00 AM sanchay09 Calculus 5 October 10th, 2013 12:21 AM enrigue Linear Algebra 1 January 26th, 2009 07:15 AM sanchay09 Algebra 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      