Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a triangular matrix L and a diagonal matrix D, what can be said about the singular values of the product D*L ?

Precisely, is it possible to express the singular values of D*L as function of the diagonals of D and the singular values of L ?

Any help would be appreciated. Thanks in advance.

share|improve this question
    
The title and body don't match. –  joriki Dec 14 '12 at 10:02
1  
No research effort whatsoever - you're more likely to get help if you have a crack yourself first. –  Epictetus Dec 14 '12 at 10:04
add comment

1 Answer

To state the obvious, you definitely cannot express the singular values of $DL$ as a function of $D$ and the singular values of $L$. For example, let $D=\mathrm{diag}(2,1), L_1 = \mathrm{diag}\left(\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}\right)$ and $L_2=\begin{pmatrix}1&1\\0&1\end{pmatrix}$. The matrices have identical singular values, but $DL_1$ and $DL_2$ haven't any singular value in common.

There are some bounds for the singular values of $DL$ based on the singular values of $D$ and $L$. For example, when both $D$ and $L$ are $n\times n$, we have (cf. Horn and Johnson, Matrix Analysis, p.423) $$ \sigma^\downarrow_{i+j-1}(DL)\le\sigma^\downarrow_i(D)\sigma^\downarrow_j(L), \quad i+j\le n+1, $$ where $\sigma^\downarrow_i(X)$ denotes the $i$-th largest singular value of a matrix $X$. However, to my understanding, preconditioners are usually chosen based on the actual matrices being preconditioned. Without knowing the connection between $D$ and $L$, even if we can obtain some general bounds for the singular values of $DL$, I doubt they are useful for preconditioning's sake.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.