- #1
mathmari
Gold Member
MHB
- 5,049
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Hey!
Let $A=L^TDL$ be the Cholesky decomposition of a symmetric matrix, at which the left upper triangular $L$ hat only $1$ on the diagonal and $D$ is a diagonal matrix with positiv elements on the diagonal.
I want to show that such a decomposition exists if and only if $A$ is positive definite.
Could you give me a hint how we could show that?
If we suppose that $A=L^TDL$ is the Cholesky decomposition then if $A$ positiv definite the diagonal elements must be positiv and so the elements of $D$ are positive, or not?
![Smiling face with smiling eyes :blush: 😊](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f60a.png)
Let $A=L^TDL$ be the Cholesky decomposition of a symmetric matrix, at which the left upper triangular $L$ hat only $1$ on the diagonal and $D$ is a diagonal matrix with positiv elements on the diagonal.
I want to show that such a decomposition exists if and only if $A$ is positive definite.
Could you give me a hint how we could show that?
![Thinking face :thinking: 🤔](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f914.png)
If we suppose that $A=L^TDL$ is the Cholesky decomposition then if $A$ positiv definite the diagonal elements must be positiv and so the elements of $D$ are positive, or not?