91Ó°ÊÓ

In mathematics and particularly in linear algebra, a matrix is called a positive definite matrix if it satisfies certain conditions. A square matrix is positive definite if for any non-zero column vector \mathbf{x}\, the scalar \mathbf{x}^T A \mathbf{x}\ is always greater than zero. Here, \mathbf{x}^T\ represents the transpose of \mathbf{x}\, and A is our matrix.
A positive definite matrix has several important properties:

• All its eigenvalues are positive.
• The matrix is invertible.
• For Cholesky factorization, the matrix must be positive definite.

In our given example, the matrix A is:
\[ \begin{bmatrix} 1 & 2 & 1 \ 2 & 8 & 10 \ 1 & 10 & 18 \end{bmatrix} \]We can check that it is positive definite, satisfying the conditions needed for Cholesky factorization.

A lower triangular matrix is a special kind of square matrix where all the elements above the main diagonal are zero. This means for a matrix L:
\[ L = \begin{bmatrix} l_{11} & 0 & 0 \ l_{21} & l_{22} & 0 \ l_{31} & l_{32} & l_{33} \end{bmatrix} \]
The elements \l_{ij} = 0\ if \i < j\, implying that everything above the diagonal is zero, while the elements on and below the diagonal can be non-zero. Lower triangular matrices are essential in matrix decompositions like the Cholesky factorization. In our exercise, we need to find a lower triangular matrix L such that:
\[ A = LL^T \]This matrix L will help us decompose the given positive definite matrix A, presenting a simpler form to work with.

Matrix decomposition involves breaking down a matrix into a product of simpler matrices, making complex matrix operations easier to handle. One common form of matrix decomposition is the **Cholesky factorization**. Cholesky factorization specifically applies to positive definite matrices and decomposes them into a product of a lower triangular matrix and its transpose:
\[ A = LL^T \]
Here's how it works for our given matrix:
Firstly, set up the matrices: we are given matrix A as:
\[ \begin{bmatrix} 1 & 2 & 1 \ 2 & 8 & 10 \ 1 & 10 & 18 \end{bmatrix} \]
and we need to find a lower triangular matrix L:
\[ L = \begin{bmatrix} l_{11} & 0 & 0 \ l_{21} & l_{22} & 0 \ l_{31} & l_{32} & l_{33} \end{bmatrix} \]Next, we equate LL^T to A and solve for the elements of L. This involves solving systems of equations step by step:

• Set \l_{11}^2 = 1\ gives \l_{11} = 1\.
• Set \l_{11}l_{21} = 2\ gives \l_{21} = 2\.
• Set \l_{11}l_{31} = 1\ gives \l_{31} = 1\.

And continue with the remaining elements. Finally, we find:
\[ L = \begin{bmatrix} 1 & 0 & 0 \ 2 & 2 & 0 \ 1 & 4 & 1 \end{bmatrix} \].
Verifying by checking that \(LL^T = A\) further ensures the correctness of our decomposition.