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I am studying for a exam and I thought about practicing the Cholesky decomposition.

If a matrix $A = A^{T}$ , the main diagonal of $A$ has only positive elements and in every row the absolute value of the element in the main diagonal $>$ the sum of the absolute values of the other elements in the same row then we can decompose $A$ to $U^{T}U$ where $U$ is a upper matrix.Well I tried solving a example on my own:

and then decide to check its validity by putting it in the Symbolab matrix calculator but I dont get the same results.

Help appreciated!

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  • $\begingroup$ There is a term for the property "in every row the absolute value of the element in the main diagonal > the sum of the absolute values of the other elements in the same row" which is: "strictly diagonally dominant" $\endgroup$
    – Jean Marie
    Jun 13, 2022 at 14:09

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Click on this link: A symmetric, diagonally dominant matrix A with real positive diagonal entries is positive definite.

This property shows that such a matrix possesses a Cholesky decomposition $U^TU$ with $U$ upper triangular.


Edit: I have spotted the error in your particular example: the constraint $bc+ed=0$ (coming from line 3 times column 2) should be $bc+ed=\color{red}{1}$.

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  • $\begingroup$ Any comment ?... $\endgroup$
    – Jean Marie
    Jun 13, 2022 at 23:22

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