October 31st, 2018, 07:54 PM  #1 
Member Joined: Feb 2018 From: Canada Posts: 42 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 positivedefinite. 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, 06:29 PM  #2 
Member Joined: Feb 2018 From: Canada Posts: 42 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 positivedefinite 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 positivedefinite 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 positivedefinite. 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 06:31 PM. 
December 2nd, 2018, 08:36 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 531 Thanks: 304 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. 

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