0
$\begingroup$

$A$ is symmetric, $L$ is lower triangular, $U$ is upper triangular. Maybe I am missing some fundamental insight here - but how can we be sure that there is always a $D$ such that $DL^T=U$?

In practice it seems to work, but I'm not sure how to argue that $D$ exists in general.

$\endgroup$
  • 1
    $\begingroup$ Does $L$ in $DL^T$ differ from $L$ in $LU$ or $A$ is symmetric? $\endgroup$ – A.Γ. Sep 19 '15 at 17:43
  • $\begingroup$ may I know what $U$ is? $\endgroup$ – R.N Sep 19 '15 at 17:49
0
$\begingroup$

Let $A=A^T$ and you have $A=LU$ with a lower-triangular invertible $L$ and an upper-triangular (echelon) $U$. Then on the one hand, the matrix $L^{-1}AL^{-T}$ is symmetric, and on the other hand, $L^{-1}AL^{-T}=UL^{-T}$ where the RHS is a product of two upper-triangular matrices, thus, an upper-triangular. The only upper-triangular matrix that is symmetric is diagonal, i.e. $UL^{-T}=D$, which gives $U=DL^T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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