I' ve been reading the review Ulrich Schollwöck: The density-matrix renormalization group in the age of
matrix product states (arXiv link)
and encountered with a question about the so called 'reduced basis transformation' on page 125 (journal page number; page 39 in the arXiv preprint).
To summarize, C is a square matrix with SVD decomposition $C=U^\dagger SV$，$C^\prime$ is another matrix that can be written as
\begin{equation}
C^\prime=P^\dagger
\left [ \begin{matrix}
S& 0&0 \\
0& S &0\\
0&0&S
\end{matrix} \right]Q
\end{equation}
where $P^\dagger P=I$ and $Q^\dagger Q=I$, but $P P^\dagger,QQ^\dagger\neq I$, it is said that the largest singular value of $C^\prime$ is smaller than that of C, because this transformation corresponds to a reduced basis transformation into orthogonal subspaces.
My question is what is reduced basis transformation and how is it related to the singular value.
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$\begingroup$ @Norbert This question is about linear algebra definitions, not computational physics. $\endgroup$– Kyle KanosCommented May 31, 2023 at 19:28
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$\begingroup$ @KyleKanos The way I read the question it is about understanding one step in the explanation of the DMRG algorithm in a review paper about the DMRG algorithm. People searching for answers on the DMRG algorithm might be searching for this, so tagging it accordingly will help them find it. (One might read it as a question about linear algebra definitions, but then my answer is completely off -- I think what the OP wants to know is why the claim made in the review is correct, and this type of relation shows up very specifically in this context.) $\endgroup$– Norbert SchuchCommented May 31, 2023 at 19:47
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$\begingroup$ @KyleKanos Otherwise, you might as well argue that a question asking why Metropolis updates preserve detailed balance is not about Monte Carlo methods, but about some arithmetics. -- Similarly, if you think it is about linear algebra and nothing else, the tags "spin-chains" and "tensor-network" ought to be removed as well, and it should be migrated to math. $\endgroup$– Norbert SchuchCommented May 31, 2023 at 19:48
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$\begingroup$ @Norbert It's asking about specific terms found in a paper (what is "reduced basis transformation" and "how is it (reduced basis transform) related to SVD"). There's nothing "computational" about definitions. I would expect users to be searching for renormalization, tensor-network, linear-algebra, quantum-information, spin-models, etc over comp-phys. $\endgroup$– Kyle KanosCommented May 31, 2023 at 20:33
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$\begingroup$ You also have enough rep to create a tag, it might be more beneficial to create a dmrg tag and allow users to filter over that, if that is your concern. $\endgroup$– Kyle KanosCommented May 31, 2023 at 20:36
1 Answer
First, note that $P^\dagger P =I$ means $P$ is an isometry, and the same for $Q$.
Next, note that $\mathrm{max\,eig}(P^\dagger MP)\le \mathrm{max\,eig}(M)$, for any hermitian matrix $M$ and any isometry $P$ (this is a consequence of the variational characterization of the largest eigenvalue, i.e. as the maximum of $\langle \phi\vert M\vert\phi\rangle$).
Let me denote the matrix with the several $S$ in $C'$ by $\hat S$.
Then, we use that the largest singular value of $C'$ squared equals \begin{align} \mathrm{max\,eig}(C'^\dagger C') &= \mathrm{max\,eig}(P^\dagger \hat S Q Q^\dagger \hat S P)\\ & \le \mathrm{max\,eig}(\hat S Q Q^\dagger \hat S)\\ & \stackrel{(*)}{\le} \mathrm{max\,eig}(Q^\dagger \hat S^2 Q)\\ &\le \mathrm{max\,eig}(\hat S^2)\ , \end{align} which is precisely the largest singular value of $C$ squared and thus completes the proof.
(Note that in (*), we have used that the non-zero spectrum is cyclic (i.e. $AB$ and $BA$ have the same non-zero spectrum), and thus, $\mathrm{max\,eig}(AB)=\mathrm{max\,eig}(BA)$.)
