@SanchayanDutta ah, so either they changed the wording since then, or the wording in the close queue is different than the one that actually ends up in the closed post
by the way, I always find it puzzling when there are closed questions without upvotes with multiple upvoted answers. Like, what's the point in that?
also, that guy really likes diagonalising matrices lol
by the way, I actually think one can answer that question in much more generality. Consider the problem of, given an Hermitian (or symmetric if real) $B$, asking for the $A$ such that $B=A^\dagger A$. It might not be immediately obvious, but this problem is essentially equivalent to that of finding a purification for a given state $\rho$, which can be solved in full generality thinking in terms of the SVD of $A$
Anonymous
9:28 AM
@glS Uh, those old-post closures are generally done by the CRUDE room fellows. If I recall correctly, there was an active attempt to close all old homework-y questions even if they had answers, to urge people to provide more context and attempts in their questions and to prevent users from gaining reputation by answering low-hanging fruits. To be fair though, I hadn't put much effort into writing that question!
When we have a symmetric matrix $A = LL^*$, we can obtain L using Cholesky decomposition of $A$ ($L^*$ is $L$ transposed).
Can anyone tell me how we can get this same $L$ using SVD or Eigen decomposition?
Thank you.
@glS If I remember correctly, the $A$ isn't unique? If we follow the straightforward algorithms of the two decompositions then Cholesky won't return the same $A$ as SVD (?), though it's possible to go from one to another
the gist is that, given $B$ (you have to assume it's positive for $B=A^\dagger A$ to be possible) with eigendecomposition $B=\sum_k p_k |u_k\rangle\!\langle u_k|$ where $p_k\ge0$, the set of $A$ such that $B=A^\dagger A$ is the set of matrices whose SVD reads $A=\sum_k \sqrt{p_k}|v_k\rangle\!\langle u_k|$ (that is, the set of matrices of this form for arbitrary orthonormal bases $\{|v_k\rangle\}$
then if you think of $A$ as a state by writing $\langle u_k|\mapsto|u_k\rangle$, you get that $A$ is a purification of the "state" $B$, so the problems are one and the same
@SanchayanDutta it might be unique in the case of the Cholesky decomposition, I don't actually know that one very well so I'm not sure. But that's a different problem because you also require lower-triangular/upper-triangular matrices
in general $A$ is most definitely not unique though. For example (using bra-ket notation for conciseness) $A=|1\rangle\!\langle 1|+2|2\rangle\!\langle 2|$ gives you the same as $A=|1123\rangle\!\langle 1|+2|251\rangle\!\langle 2|$
or more concretely if you prefer $\begin{pmatrix}1&0\\0&2\end{pmatrix}$ gives the same as $\begin{pmatrix}0&0\\1&0\\0&0\\0&2\end{pmatrix}$
Anonymous
Right, makes sense. I didn't think of the SVD (\purification) method back. That would probably give the general algorithm for calculating all possible $A$ such that $AA^{\dagger} = B$ where $B$ is Hermitian?
Anonymous
9:58 AM
> Eq. (B) makes it very clear what the possible purifications of 𝜌 are: the freedom is all and only in the choice of an orthonormal set of rank(𝜌) vectors from some arbitrary ancillary space (with the only caveat that this space needs to be large enough to accommodate this number of orthogonal elements).
Anonymous
So yeah, I guess we just need to exhaustively search all the orthonormal basis sets to get all the possible A's
@SanchayanDutta well, yes, the general form is the one I wrote for arbitrary sets of orthonormal vectors $|v_k\rangle$. If you want you can parametrise such sets in terms of angles and phases, but whether you want to do that just depends on what the problem you are trying to solve is
When we have a symmetric matrix $A = LL^*$, we can obtain L using Cholesky decomposition of $A$ ($L^*$ is $L$ transposed).
Can anyone tell me how we can get this same $L$ using SVD or Eigen decomposition?
Thank you.
