Can a Symmetric Positive Definite Matrix Always Be LU Factorized?

[buzzmath]

Homework Statement

A is an nxn symmetric and positive definate matrix. Show that A admits a factorization A=LU
In other words, no zero pivot is encountered during the elimination process.

Homework Equations

cholesky factorization

The Attempt at a Solution

I think all I have to show is that after 1 step in G.E. a(1,1) does not equal 0 and that the first row is the same as the first row in A and the first column is all zeros except for a(1,1). then show that the remaining (n-1)x(n-1) matrix is symmetric and positive definate. I showed that a(1,1) is not zero but I don't know how to prove the (n-1)x(n-1) matrix is positive definate and symmetric.
any help or suggestions?

 

[mighty2000]

Hello, thank you for your post! You are on the right track with showing that a(1,1) is not equal to 0 in the first step of Gaussian elimination. This is because if a(1,1) is equal to 0, then the matrix A is not positive definite.

To show that the remaining (n-1)x(n-1) matrix is symmetric and positive definite, you can use the Cholesky factorization. This factorization states that a symmetric positive definite matrix A can be written as A = LL^T, where L is a lower triangular matrix with positive diagonal elements.

Using this factorization, you can show that the remaining (n-1)x(n-1) matrix is also positive definite and symmetric by showing that it can be written as L'L', where L' is also a lower triangular matrix with positive diagonal elements.

I hope this helps! Let me know if you need any further clarification. Good luck with your proof!