Is a symmetric matrix positive definite iff $D$ in its LDU decomposition is positive definite? Given
$$A=LDU$$
 where


*

*$A$ is a real symmetric matrix

*$L$ is a lower unitriangular matrix

*$D$ is a diagonal matrix

*$U$ is an upper unitriangular matrix


can we say that
$$A>0 \iff D>0$$
?
Edit:
My thinking is that $(LD^{1/2})(D^{1/2}U)$ is (probably?) the Cholesky decomposition, and $D^{1/2}$ exists iff $D>0$.

First of all, if $A = LDU$ is symmetric, we must have $U=L^T$ by uniqueness of the decomposition ($LDU = A=A^T = U^T D L^T$). so then $A=LDL^T$.
So if $D>0$ then by for all $x\neq 0$ denoting $y=L^Tx$ (why is $y \neq 0$?) yields
$$x^T A x = x^T L D L^T x = (L^T x)^T D L^T x = y^T D y > 0$$
And if $A>0$ then for all $x \neq 0 $ denoting $y = (L^T)^{-1}x $ yields
$$x^T D x = (L^T y)^T D (L^T y)=y^T LDL^T y = y^T A y > 0$$
