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$A$ is symmetric positive definite and $A = LDL^T$, where $L$ is unit lower triangular and $D$ is diagonal. I want to prove that the main-diagonal entries of $D$ are all positive.

I have tried $\det(A)>0 \Rightarrow \det(LDL^T)>0$. Since $\det(L)=\det(L^T)=1$, $\det(D)$ must be positive. But, that doesn't mean all of the main diagonal entries of D are positive. Should I be using a different property?

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    If $X$ is non-singular and $A$ is PD then $X^T A X$ is PD. Take $X = L^{-1}$.2012-02-25
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    @J.D. Do you mean let $X=(L^{-1})^T$? So that $X^TAX=D$2012-02-27
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    Yes. That's what I meant.2012-02-27

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