# Polar decomposition - expression of the isometry

Suppose $$T \in \mathcal{L(V)}$$ is invertible and has singular value decomposition given by $$T(v)=s_1\langle v,e_1\rangle f_1+\cdots+s_n\langle v,e_n\rangle f_n$$ for every $$v \in \mathcal{V}$$, where $$s_1, \dots, s_n$$ are the singular values of $$T$$ and $$e_1, \dots, e_n$$ and $$f_1,\dots,fn$$ are orthonormal bases of $$\mathcal{V}$$. I know that for every $$v \in \mathcal{V}$$ $$T^*(v)=s_1\langle v,f_1\rangle e_1+\cdots+s_n\langle v,f_n\rangle e_n \\ T^*T(v)=s_1^2\langle v,e_1\rangle e_1+\cdots+s_n^2\langle v,e_n\rangle e_n; \\ \sqrt{T^*T}(v)=s_1\langle v,e_1\rangle e_1+\cdots+s_n\langle v,e_n\rangle e_n; \\ T^{-1}(v)=\frac{\langle v,f_1\rangle}{s_1} e_1+\cdots+\frac{\langle v,f_n\rangle}{s_n} e_n.$$ I also know, from the polar decomposition, that there is a isometry $$S \in \mathcal{L(V)}$$ such that $$T=S \sqrt{T^∗T}$$. Based on the expressions above, how can I deduce a mathematical expression for $$S$$?

• You can solve the equation $T = S\sqrt{T^*T}$ to get $S = T(T^*T)^{-1/2}$. Commented Oct 30, 2021 at 22:23

First of all, solve the equation $$T = S\sqrt{T^*T}$$ to get $$S = T(T^*T)^{-1/2}$$. Verify that the inverse of $$\sqrt{T^*T}$$ is given by $$(T^*T)^{-1/2}(v) = \sum_i s_i^{-1} \langle v,e_i\rangle e_i.$$ Next, compute the composition \begin{align} T(T^*T)^{-1/2}(v) &= T \sum_i s_i^{-1} \langle v,e_i\rangle e_i \\ & = \sum_j s_j \left \langle \sum_i s_i^{-1} \langle v,e_i\rangle e_i,e_j \right \rangle f_j \\ & = \sum_{i,j} s_i^{-1}s_j \langle v,e_i \rangle \langle e_i,e_j \rangle f_j \\ & = \sum_{i} s_i^{-1}s_i \langle v,e_i \rangle f_i = \sum_{i} \langle v,e_i \rangle f_i. \end{align}