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The Wikipedia article on the polar decomposition states that, for any matrix $A \in \mathbb{R}^{m \times n}$, the polar decomposition is defined as $A = UP$ where $U \in \mathbb{R}^{n \times m}$ and $P \in \mathbb{R}^{m \times m}$, it then states that we can get the polar decomposition by using the SVD and setting $P = V\Sigma V^*$ and $U = WV^*$ but SVD is defined as $A = W\Sigma V^*$ where $W \in \mathbb{R}^{m\times m}$, $\Sigma \in \mathbb{R}^{m\times n}$ and $V \in \mathbb{R}^{n\times n}$.

As far as I can see the dimensions don't add up for the multiplications in $P$ and $U$ or what am I missing ($P(m\times m) = V(n\times n)\cdot \Sigma(m\times n) \cdot V^*(n\times n)$ and $U(n\times m) = W(m\times m)\cdot V^*(n\times n)$)?

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  • $\begingroup$ Your definition of $A$ is $A \in \mathbb{R}^{m \times n}$, and $A = UP$ should be defined with $U \in \mathbb{R}^{m \times n}$ and $P \in \mathbb{R}^{n \times n}$. $\endgroup$
    – Tab1e
    Apr 25, 2022 at 12:57
  • $\begingroup$ And I think that another definition of SVD is being used in this context. $\endgroup$
    – Tab1e
    Apr 25, 2022 at 12:59
  • $\begingroup$ but then $UP \in \mathbb{R}^{m \times m}$ isn't it? $\endgroup$ Apr 25, 2022 at 13:00
  • $\begingroup$ What do you mean by $UP \in \mathbb{R}^{m \times m}\ $? Are you assuming that $A$ is a square matrix? $\endgroup$
    – Tab1e
    Apr 25, 2022 at 13:01
  • $\begingroup$ en.m.wikipedia.org/wiki/Singular_value_decomposition this article is linked $\endgroup$ Apr 25, 2022 at 13:02

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There are two sets of definitions with SVD. For $A \in M_{n\times m}$, one is $$ A = W \cdot \Sigma \cdot V^T $$ with $W \in M_{n \times l},\ \Sigma \in M_{l \times l},\ V \in M_{m \times l}$ where $l = \min\{m,n\}$. And another one is $$ A = W \cdot \Sigma \cdot V^T $$ with $W \in M_{n \times n},\ \Sigma \in M_{n \times m},\ V \in M_{m \times m}$.

To find the connection between SVD and the polar decomposition, remember that for any matrix $A$, the matrix $P$ is uniquely determined by $(A^TA)^{1/2}$.

For the first definition, in the original post, one simply has $$ A = W \Sigma V^T =(W V^T)(V \Sigma V^T)= UP $$ However, for the second definition, the computation is tricker as the matrix $\Sigma$ is not a square matrix. First compute $$ A^TA = V \Sigma^T \Sigma V^T = V D V^T $$ where $D = \Sigma^T \Sigma = diag(\sigma_1^2,\dots,\sigma_r^2,0,\dots,0) \in \mathbb{R}^{n \times n}$ with $Rank(A) = r$. Again, the matrix $P$ is defined as $$ P = V D^{1/2} V^T $$ and you should try to finish the rest of the formulation.

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  • $\begingroup$ Ah I see I've totally overlooked the thin SVD formulation, this feels much more natural if you first look at the polar decomposition. I'm still not 100% sure on how to construct the $U$ matrix but my best guess would be to set $U = WI_{m\times n}V^*$ since in the other formulation the remaining vectors would map into the nullspace $\endgroup$ Apr 25, 2022 at 17:13

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