# Deriving SVD from polar decomposition

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

• 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 $P \in \mathbb{R}^{n \times n}$. Apr 25, 2022 at 12:57
• And I think that another definition of SVD is being used in this context. Apr 25, 2022 at 12:59
• but then $UP \in \mathbb{R}^{m \times m}$ isn't it? Apr 25, 2022 at 13:00
• What do you mean by $UP \in \mathbb{R}^{m \times m}\$? Are you assuming that $A$ is a square matrix? Apr 25, 2022 at 13:01
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.
• 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 Apr 25, 2022 at 17:13