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I have a matrix in LDLT $X=LDL^T$ form, where all nonzero elements in L (i.e. diagonal and below), are $1$. So only D matters (which is diagonal). What can we tell about the eigenvalues/vectors of this matrix?

It is possible to assume for simplicity that $X$ is positive definite, i.e. D's diagonal is all positive.

Troy McClure
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  • Did this question just change completely? – copper.hat Dec 07 '12 at 20:24
  • yes, but it's completely equivalent to the previous one! – Troy McClure Dec 07 '12 at 20:34
  • I may be missing something, but in the previous question, $U$ had ones above and on the diagonal from bottom left to upper right instead of the usual triangular? – copper.hat Dec 07 '12 at 22:17
  • this is the conjugate case, of the inverse $X^-1$. invert a matrix with L having all 1 as described here, and you'll get 1 in the diagonal and -1 in the subdiagonal, and zero elsewhere. – Troy McClure Dec 07 '12 at 22:58
  • i.e., the inverse of the X here is tridiagonal – Troy McClure Dec 07 '12 at 23:01
  • Not sure if this is the sort of thing you are looking for, but Sylvester's law of inertia shows that the number of negative (zero, positive, resp.) eigenvalues of $X$ is the same as the number of negative (zero, positive, resp.) diagonal values of $D$. (Of course, this only requires $L$ to be invertible, so presumably one can say more here.) – copper.hat Dec 07 '12 at 23:25
  • We can assume full rank and all positive eigenvalues, i.e. we can assume $X$ is symmetric and positive definite. Also, L is invertible in our case. The problem is that its vectors are only independent, not orthogonal. But as expected I need the eigenvalues themselves, not their count. Thanks!!! – Troy McClure Dec 08 '12 at 01:17

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