Question

Consider the matrix \( A = \begin{pmatrix} 9 & 15 \\ 15 & 50 \end{pmatrix} \). The matrix A is decomposed using Cholesky decomposition. Determine the value of \(l_{22}\).

Hint:

For a 2x2 matrix \(A = \begin{pmatrix} a & b
b & c \end{pmatrix}\), the Cholesky elements can be found directly: \(l_{11} = \sqrt{a}\), \(l_{21} = b/l_{11}\), and \(l_{22} = \sqrt{c - l_{21}^2}\).

Answer 1

Step 1: Understanding Cholesky Decomposition:
Cholesky decomposition is a method for decomposing a symmetric, positive-definite matrix \(A\) into the product of a lower triangular matrix \(L\) and its conjugate transpose \(L^T\). The equation is \(A = LL^T\). The diagonal elements of L are required to be real and positive.
Step 2: Setting up the Matrices:
Given matrix: \( A = \begin{pmatrix} 9 & 15 \\ 15 & 50 \end{pmatrix} \). Let the lower triangular matrix be \( L = \begin{pmatrix} l_{11} & 0 \\ l_{21} & l_{22} \end{pmatrix} \). Its transpose is \( L^T = \begin{pmatrix} l_{11} & l_{21} \\ 0 & l_{22} \end{pmatrix} \). The product \(LL^T\) is: \[ LL^T = \begin{pmatrix} l_{11} & 0 \\ l_{21} & l_{22} \end{pmatrix} \begin{pmatrix} l_{11} & l_{21} \\ 0 & l_{22} \end{pmatrix} = \begin{pmatrix} l_{11}^2 & l_{11}l_{21} \\ l_{21}l_{11} & l_{21}^2 + l_{22}^2 \end{pmatrix} \] Step 3: Solving for the Elements of L:
We equate the elements of \(A\) with the elements of \(LL^T\): \[ \begin{pmatrix} 9 & 15 \\ 15 & 50 \end{pmatrix} = \begin{pmatrix} l_{11}^2 & l_{11}l_{21} \\ l_{21}l_{11} & l_{21}^2 + l_{22}^2 \end{pmatrix} \]

1. From the (1,1) element: \( l_{11}^2 = 9 \implies l_{11} = \sqrt{9} = 3 \) (we take the positive root).
2. From the (2,1) element: \( l_{21}l_{11} = 15 \implies l_{21} \times 3 = 15 \implies l_{21} = 5 \).
3. From the (2,2) element: \( l_{21}^2 + l_{22}^2 = 50 \). Substitute the value of \(l_{21}\): \[ (5)^2 + l_{22}^2 = 50 \] \[ 25 + l_{22}^2 = 50 \] \[ l_{22}^2 = 25 \implies l_{22} = \sqrt{25} = 5 \]

Step 4: Final Answer:
The value of \(l_{22}\) is 5.