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 $U(n\times m) = W(m\times m)\cdot V^*(n\times n)$)?
Best Answer
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.