# Cholesky decomposition of a strictly diagonaly dominant symmetric matrix

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!

• 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" Jun 13, 2022 at 14:09

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 $$bc+ed=\color{red}{1}$$.