$A = LDL^T \Rightarrow $all of the main diagonal entries of $D$ are positive? $A$ 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?
 A: Let $e_j$ the $j$-th vector of the canonical basis. Since $L$ is invertible we are allowed to write $(L^t)^{-1}e_j$ and since $A$ is positive definite $$0<((L^{-1})^te_j)^tA(L^{-1})^te_j=e_j^tL^{-1}A(L^{-1})^{t}e_j=e_j^tDe_j=D_{jj}$$
and we are done.
A: There are probably many ways of doing it. Here is one:
Note first that $A$ admits a square root, i.e. there is a symmetric positive definite matrix $A^{1/2}$ such that $(A^{1/2})^2=A$. This is done by writing $A$, via the Spectral Theorem, as 
$$
A=\sum_{j=1}^n \lambda_j P_j,
$$
where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ counting multiplicities, and $P_1,\ldots,P_n$ are pairwise orthogonal projections of rank one (i.e. the projections onto the corresponding eigenspaces). Then define
$$
A^{1/2}=\sum_{j=1}^n \lambda_j^{1/2} P_j.
$$
Now, since $L$ is invertible, we can write 
$$
D=MAM^T,
$$
where $M=L^{-1}$. 
Considering the canonical basis $\{e_1,\ldots,e_n\}$, we have
$$
D_{kk}=\langle De_k,e_k\rangle = \langle MAM^Te_k,e_k\rangle
=\langle A^{1/2}M^Te_k, A^{1/2}M^Te_k\rangle\geq0.
$$
But $D$ is invertible, so $D_{kk}\ne0$, and so $D_{kk}>0$ for all $k=1.\ldots,n$. 
