Is the Matrix Symmetric Positive Definite for Cholesky Decomposition?

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SUMMARY

The discussion centers on determining whether the matrix A, defined as [[1, 2, 0, 0], [2, 9, 1, 0], [0, 1, 9, 4], [0, 0, 4, 3]], is symmetric positive definite for the purpose of applying the Cholesky decomposition method. The key condition for a matrix to be symmetric positive definite is that the maximum off-diagonal element must be less than the maximum diagonal element. In this case, the maximum diagonal element a_44 = 3 is less than the maximum off-diagonal element a_43 = 4, indicating that the matrix does not satisfy the necessary condition for Cholesky decomposition.

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fonseh
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Homework Statement


Here's the question :
1x1+ 2x2 +0x3 + 0x4 = 1
2x1+ 9x2 +1x3 + 0x4 = 6
0x1+ 1x2 +9x3 + 4x4 = 2
0x1+ 0x2 +4x3 + 3x4 = 8

I' m asked to solve this question using Choelsky method ( We need the symmetric positive definite matrix when we are using this method)

Homework Equations

The Attempt at a Solution


matrix A = $$\begin{bmatrix}
1 & 2& 0 & 0 \\
2 & 9 & 1 & 0 \\\
0 & 1 & 9 & 4 \\
0 & 0 & 4 & 3
\end {bmatrix} $$


the book stated that for the positive symmetric matrix , we need to ensure that max a_kj less than max a_ii ,
But , in this example , i found that the a_44 which is 3 is less than a_43 which is 4 ... So , how could this be symmetric positive definite matrix ?
How is it possible to solve using Choelsky method ?
 

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fonseh said:
I' m asked to solve this question using Choelsky method
The guy is called Cholesky.

But it's clear what you are being asked, so go to work and see if and where it goes wrong !
 
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question solved
 

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