**Suppose that $S$ is an isometry, and $T$ is an operator on a vector space V. Show that singular values of T and ST coincide.
Suppose $S$ is isometry and suppose that $T\in L(V)$ and T has singular values $ s_1,...,s_n $. By singular decomposition, there exists orthonormal basis $ (e_1,...,e_n)$ and $(f_1,...,f_n)$ of V.
Then, $ Tv = s_1 \langle v,e_n \rangle f_1 + ... + s_n \langle v,e_n \rangle f_n$ Then $ (ST)v = s_1 \langle v,e_n \rangle S f_1 + ... + s_n \langle v,e_n \rangle S f_n$ $ = s_1 \langle v ,e_1 \rangle f_1 + ... + s_n \langle v, e_n \rangle f_n = T v $
Hence, $ Tv = ST v $ and their singular values coincide.
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** Definition of singular values: Suppose $ T \in L(V)$. The singular values of T are the eigenvales of $ \sqrt {T^*T}$ , with each eigen value $\lambda$ repeated dim null $ \sqrt {T^*T} - \lambda I$ times**