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 ?

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.