## Spectral properties of the LDL^T matrix factorization

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Assume that a square, symmetric matrix $A$ can be factored into $A=LDL^T$ where $L$ is unit lower triangular and $D$ is diagonal. For indefinite $A$, $D$ may have $2x2$ blocks on the diagonal. How much information about the spectrum of $A$ can we obtain from $D$?

For example, it is known by Sylvester's law of matrix inertia that the inertia of $D$ is the same as that of $A$ (they have the same number of positive and negative eigenvalues). This is interesting, but I am wondering what other information is hidden in $D$.

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I was always wondering the same :) Great question.  – Alexander Chervov Dec 28 2011 at 8:18
But why you write about 2x2 blocks in indefinite case ? Usually just elements of D can  be taken negative. – Alexander Chervov Dec 28 2011 at 8:21

## 2 Answers

Well, for the closely related Cholesky factorization, there is the following:

Fast Accurate Eigenvalue Computations Using the Cholesky Factorization (1997) (by Roy Matthias), which says that the eigenvalues are very close to the squares of the diagonal elements of the Cholesky factor. (the paper is available on CiteSeer).

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I've heard similar, but there were some conditions, may be small eigs, or like that... But "squares" seems to me misprint - take A=D then they are equal. Any way... What can be the reason for relation in case "L" is not very small ? Is there some intuitive explanation ? – Alexander Chervov Dec 28 2011 at 8:23
In Cholesky, $D=I$ and the diagonal scaling is "included" in the $L$ factor, which need not have ones on its main diagonals. So that's where you get those square roots from. – Federico Poloni Dec 28 2011 at 8:57
@Federico, You mean - probably the paper mentioned above deals with LL^t ? Then it is Okay- we need squares. Just  in the question  LDL^t was mentioned which is sometimes called Cholesky or Cholesky without square roots also. In this case you do not need square roots. Any way puzzle seems to be resolved :) What about my questions ? What the reason can be to have eigs = D ?  – Alexander Chervov Dec 28 2011 at 10:05

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Not an answer, but just thinking loudly (can I say like this in English?)

Consider A=

$a ~~ b$

$b ~~ a$

Then eigenvalues are equal to a-b, a+b (it is easy to check since trace and determinant are correct).

Then L =

$1~~~~~~ 0$

$b/a~~1$

D=

$a ~~~~ 0$

$0 ~~~~ a - b^2/a$

(It is easy in 2x2 case since determinant(A) = $d_1d_2$ ).

So we see: eigenvalues are:

$a \pm b$ and elements of D are $a$ and $a-b^2/a$

Well, do they look similar ?

Stupid case when they are similar is b=0.

Another case is more interesting - take b=a - very degenerate but will be positve under small perturbation a=b+small. So in this case eigs are : $2a, 0$, and elements of D are $a$ and $0$ . So we see that the smallest number (i.e. $0$) is the same.

Probably that is the phenomena which I heard about i.e. something similar holds true for NxN matrices.

Probably paper mentioned in Igor's answer is something related.

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