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October 31st, 2018, 06:54 PM   #1
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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.
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November 30th, 2018, 05:29 PM   #2
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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.
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.
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December 2nd, 2018, 07:36 PM   #3
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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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