Suppose $A$ is an $n\times n$ symmetric positive definite matrix. We know that $A$'s leading principal minors $m_1,m_2,\dots,m_n$ are positive.
Now suppose that $A$ has LDU decomposition $A=LDU$, and let $d_1,d_2,\dots,d_n$ denote the diagonal entries of $D$. By multiplying out $LDU$, I found that $d_1 = m_1$ and $d_2 = m_2/m_1$. I then conjectured that $d_i = m_i/m_{i-1}$ generally, and confirmed this with some elbow grease for $i=3$.
Unfortunately, I just don't have enough calculational fluency with matrices to prove the general case myself. I'd love to see why it might be true, and hopefully in the process learn some calculational tools for dealing with determinants, minors, and other matrix manipulations.
Thanks in advance!
Let $A_k$ be the $k\times k$-matrix that consists of the first $k$ rows of the first $k$ columns of $A,$ $L_k$ the $k\times k$-matrix that consists of the first $k$ rows of the first $k$ columns of $L,$ $D_k$ the $k\times k$-matrix that consists of the first $k$ rows of the first $k$ columns of $D$ and $U_k$ the $k\times k$-matrix that consists of the first $k$ rows of the first $k$ columns of $U.$
It can be easily verified that $A_k = L_kD_kU_k.$ (When you perform the calculations for $LDU,$ you will notice that you only use $L_k,$ $D_k$ and $U_k$ to get the elements of $A_k,$ because of all the zeroes in $L$ and $U.$)
Therefore, using the multiplicativity of the determinant and the fact that the determinant of a triangular matrix is just the product of its diagonal elements, we have $$ m_k = \det(A_k) = \det(L_kD_kU_k) = \det(L_k)\det(D_k)\det(U_k) = 1\cdot\prod_{j=1}^k d_j \cdot 1= \prod_{j=1}^k d_j $$ from which your conjecture follows.