Reconstruct A from its reduced eigenparis

  • Thread starter Thread starter jollage
  • Start date Start date
jollage
Messages
61
Reaction score
0
Hi all,

Suppose I have a matrix A_{N\times N}. I compute its eigenmodes

A V = V \Lambda.

V, \Lambda are eigenvectors and eigenvalues of size N\times N. The eigenvalues are descending.

Now I cut off several eigenmodes (the ones having small value), it becomes

A V_{N \times n} = V_{N \times n} \Lambda_{n \times n}.

What I want is to reconstruct A from V_{N \times n}, \Lambda_{n \times n}.

It seems that in Matlab, if I just modify the above equation

A = V_{N \times n} \Lambda_{n \times n}V^{-1}_{N \times n},

it will not work.

So my question is: how can I reliably reconstruct the original matrix by using its reduced eigenmodes? Thanks!
 
Last edited:
Physics news on Phys.org
I don't think you can do this for an arbitrary matrix ##A##. It's not clear what you mean by ##V_{N\times m}^{-1}## for a non-square matrix ##V_{N\times m}##.

You can do something similar with the singular value decomposition ##A = U\Sigma V^T## where ##U## and ##V## are unitary matrices.

You can also do what you want if ##A## is Hermitian, because ##V^{-1} = V^T##.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top