# If $A$ can be factored as $LU$, is there always a diagonal matrix $D$ such that $DL^T=U$?

$A$ 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.

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

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 $U=DL^T$.