Let $T\in L(\mathbb{R}^{n},\mathbb{R}^{m})$ with $n\leq m$. If $R\in L(\mathbb{R}^{n})$, how to prove that $|| T\circ R|| =||T|||R|||$. Here i am using two facts: Polar decomposition
When i have $V,W$ Hilbert spaces and $L:V\to W$ linear application then
i) if $dim V\leq dim W$ then exist $S\in Sym (n)$ and $O\in O(V,W)$ such that $L=O\circ S$
ii) If $dim W\leq dim V$ then exist $S\in Sym (n)$ and $O\in O(V,W)$ such that $L=S\circ O^{*}$
In this case, when $L=O\circ S$ or $L=S\circ O^{*}$ then $||L||:=|Det S|$.
My attemp , using polar decomposition for $T=O\circ S$ imply
$||T\circ R||=||O\circ S\circ R||=|det ( S\circ R)|$ because $R$ is inyective, and using that $ O\circ O^{*}=I$, and (Im not sure if) the las equality if take $ |det ( S\circ R)|=|det ( S\circ O\circ O^{*}\circ R)|$ can to help in something, or maybe take $L_1:=S\circ O$ and $L_2=O^{*}\circ R$, Can somebody help me to continued please, or give one hint, other way to prove that, tahnk you!!!
Note (or verify using the polar decomposition) that $\|M\| = \sqrt{\det(M^*\circ M)}$ for $M \in L(\Bbb R^n, \Bbb R^m)$ (with $n \leq m$) and note that $\det(A\circ B) = \det(A) \det(B)$ for $A,B \in L(\Bbb R^n)$. With that, we have $$ \begin{align} \|T \circ R\|^2 &= \det([T \circ R]^* \circ [T \circ R]) \\ & = \det(R^* \circ T^* \circ T \circ R) \\ & = \det(R^* \circ [T^* \circ T] \circ R) \\ & = \det(R^*) \det(T^* \circ T) \det(R) \\ & = \det(T^* \circ T)[\det(R^*)\det(R)] \\ & = \det(T^* \circ T) \det(R^* \circ R) = \|T\|^2\|R\|^2. \end{align} $$ Thus, $\|T\circ R\| = \|T\|\cdot \|R\|$.
Proof of the statement: let $M = U \circ S$ be a polar decomposition (so $U$ is orthogonal injective and $S$ is symmetric). It follows that $$ M^* \circ M = (U \circ S)^* \circ (U \circ S) = S^* \circ U^* \circ U \circ S = S \circ (U^* \circ U) \circ S = S^2. $$ It follows that $\det(M^* \circ M) = \det(S^2) = \det(S)^2$. Thus, we have $$ \sqrt{\det(M^* \circ M)} = \sqrt{\det(S)^2} = |\det(S)| = \|M\|, $$ which was what we wanted.