Polar decomposition - expression of the isometry

Question

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 $S$?

Answer 1

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 $$ \begin{align} T(T^*T)^{-1/2}(v) &= T \sum_i s_i^{-1} \langle v,e_i\rangle e_i \\ & = \sum_j s_j \left \langle \sum_i s_i^{-1} \langle v,e_i\rangle e_i,e_j \right \rangle f_j \\ & = \sum_{i,j} s_i^{-1}s_j \langle v,e_i \rangle \langle e_i,e_j \rangle f_j \\ & = \sum_{i} s_i^{-1}s_i \langle v,e_i \rangle f_i = \sum_{i} \langle v,e_i \rangle f_i. \end{align} $$