# Singular values in linear algebra

**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.

** 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**

• Are we allowed to assume that $V$ is a finite dimensional inner product space? Nov 21, 2014 at 15:40
• Ah it is not given tho. So my answer is wrong? Nov 21, 2014 at 15:42
• 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. Nov 21, 2014 at 15:44
• Note that in infinite-dimensional spaces, isometries are not generally surjective. Also, why do you have $Sf_i = f_i?$ Nov 21, 2014 at 15:46
• I did it! Please see the question again! Nov 21, 2014 at 16:32

Hint: Note that $$(ST)^*(ST) = T^*(S^*S)T$$ Verify that $S^*S = I$. From there, it suffices to apply the definition.
• Then I $(ST)^* (ST) = T^*T = T^2$. Then are we allowed to assume that T is projection operator? Thanks for your hints tho! Nov 21, 2014 at 16:38
• Why did you decide that $T^*T = T^2$??? Nov 21, 2014 at 16:39
• 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? Nov 21, 2014 at 16:40
• Yes. Since $\sqrt{T^*T}$ and $\sqrt{(ST)^*ST}$ are the same transformation, the singular values have to be the same. Nov 21, 2014 at 16:41