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A is symmetric positive definite. So A = TT', where T is lower triangular matrix whose diagonal elements are all strictly positive. Someone told me which I don't understand:

One way to express $t_{ii}^2$ is from the ith leading principal matrix of A, which is

$$ \mathbf{A}_{ii} = \left[ \begin{array}{c} \mathbf{A}_{i-1, i-1} & \mathbf{a}_{i, 1} \\ \mathbf{a'}_{i, 1} & a_{ii} \end{array} \right], $$

$t_{ii}^2 = a_{ii} - \mathbf{a'}_{i, 1} \mathbf{A^{-1}}_{i-1, i-1} \mathbf{a}_{i, 1}$ for i=2,...,n. And $t_{11}^2 = a_{11}$

Mittens
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Jackie
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  • In the Cholesky-Banachiewicz-algorithm that I have been using for almost thirty years now we never have to invert parts of the matrix $\mathbf{A}$. It is much simpler. – Kurt G. Oct 16 '22 at 15:36
  • @KurtG. I should have added more background of this question. I need the step to prove that given a symmetric positive definite matrix A, the determinant of A is less than or equal to the product of $a_{ii}$. – Jackie Oct 16 '22 at 15:42
  • $\det\big(A\big)\leq \prod_{k=1}^n a_{k,k}$ by Hadamard's Determinant Inequality – user8675309 Oct 16 '22 at 16:06

1 Answers1

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In this answer I show that the determinant of $\mathbf{A}$ equals (in your notation) $$\tag{1} \operatorname{det}\mathbf{A}=t_{11}^2...t_{nn}^2 $$ (which is not that hard to see when we use $\operatorname{det}\mathbf{A}=\operatorname{det}\mathbf{T}\operatorname{det}\mathbf{T}'=(\operatorname{det}\mathbf{T})^2\,$).

From the Cholesky-Banachiewicz-algorithm we know that $$ t_{ii}^2=a_{ii}-\sum_{k=1}^{i-1}(T_{ik})^2\le a_{ii}\,. $$ Therefore, by (1), $$ \operatorname{det}\mathbf{A}\le a_{11}...a_{nn}\,. $$

Kurt G.
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