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**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.

Please give me feedback on my answer.

Thanks in advance!

** 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**

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  • $\begingroup$ Are we allowed to assume that $V$ is a finite dimensional inner product space? $\endgroup$
    – Ben Grossmann
    Nov 21, 2014 at 15:40
  • $\begingroup$ Ah it is not given tho. So my answer is wrong? $\endgroup$
    – needhelp
    Nov 21, 2014 at 15:42
  • $\begingroup$ I'm not sure. Generally, singular values are only defined for Hilbert spaces, so I think we can assume that an inner product exists. Has your book ever used the phrase "Hilbert space"? If not, you're probably allowed to assume that the space is finite dimensional. If not, then your proof is incomplete. $\endgroup$
    – Ben Grossmann
    Nov 21, 2014 at 15:44
  • $\begingroup$ Note that in infinite-dimensional spaces, isometries are not generally surjective. Also, why do you have $Sf_i = f_i?$ $\endgroup$
    – Ben Grossmann
    Nov 21, 2014 at 15:46
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    $\begingroup$ I did it! Please see the question again! $\endgroup$
    – needhelp
    Nov 21, 2014 at 16:32

1 Answer 1

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Hint: Note that $$ (ST)^*(ST) = T^*(S^*S)T $$ Verify that $S^*S = I$. From there, it suffices to apply the definition.

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  • $\begingroup$ Then I $ (ST)^* (ST) = T^*T = T^2 $. Then are we allowed to assume that T is projection operator? Thanks for your hints tho! $\endgroup$
    – needhelp
    Nov 21, 2014 at 16:38
  • $\begingroup$ Why did you decide that $T^*T = T^2$??? $\endgroup$
    – Ben Grossmann
    Nov 21, 2014 at 16:39
  • $\begingroup$ Ahh I think I kinda got it. I just have to show what you have shown then I can do square root of $T^*T$ then apply the definition. Is it correct? $\endgroup$
    – needhelp
    Nov 21, 2014 at 16:40
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    $\begingroup$ Yes. Since $\sqrt{T^*T}$ and $\sqrt{(ST)^*ST}$ are the same transformation, the singular values have to be the same. $\endgroup$
    – Ben Grossmann
    Nov 21, 2014 at 16:41
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    $\begingroup$ Thanks a lot! The answer seems so simple now but I have struggled for hours haha. Thanks a lot and you have a good day! $\endgroup$
    – needhelp
    Nov 21, 2014 at 16:41

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