Question

Use the Cholesky separation method to determine whether the following matrices are positive definite. For each that is, determine the corresponding lower diagonal matrix L: $$ \mathrm{A}=\left(\begin{array}{ccc} 2 & 1 & 3 \\ 1 & 3 & -1 \\ 3 & -1 & 1 \end{array}\right), \quad \mathrm{B}=\left(\begin{array}{ccc} 5 & 0 & \sqrt{3} \\ 0 & 3 & 0 \\ \sqrt{3} & 0 & 3 \end{array}\right) $$

Short answer

Matrix A is not positive definite. Matrix B is positive definite; the corresponding lower diagonal matrix L is \( \begin{pmatrix} \sqrt{5} & 0 & 0 \ 0 & \sqrt{3} & 0 \ \sqrt{3}/\sqrt{5} & 0 & \sqrt{12/5} \end{pmatrix} \).

Step-by-step solution

- Check if Matrix A is Symmetric

To apply the Cholesky decomposition, the matrix must be symmetric. Check if matrix A is symmetric by comparing it to its transpose. \[ \text{If} \, A = A^T, \, \text{then} \, A \, \text{is symmetric.} \] Matrix A: \[ A = \begin{pmatrix} 2 & 1 & 3 \ 1 & 3 & -1 \ 3 & -1 & 1 \end{pmatrix} \] Its transpose: \[ A^T = \begin{pmatrix} 2 & 1 & 3 \ 1 & 3 & -1 \ 3 & -1 & 1 \end{pmatrix} \] Since \( A = A^T \), matrix A is symmetric.

- Compute the Leading Principal Minors of Matrix A

Compute the leading principal minors to check for positive definiteness. The leading principal minors should be positive: \( M_1 = 2 \), \( M_2 = \det \begin{pmatrix} 2 & 1 \ 1 & 3 \end{pmatrix} = 2 \cdot 3 - 1 \cdot 1 = 5 \), \( M_3 = \det \begin{pmatrix} 2 & 1 & 3 \ 1 & 3 & -1 \ 3 & -1 & 1 \end{pmatrix} \). For \( M_3 \): \[ \begin{vmatrix} 2 & 1 & 3 \ 1 & 3 & -1 \ 3 & -1 & 1 \end{vmatrix} = 2 \begin{vmatrix} 3 & -1 \ -1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \ 3 & 1 \end{vmatrix} + 3 \begin{vmatrix} 1 & 3 \ 3 & -1 \end{vmatrix} = 2 (3\cdot1 - (-1)\cdot(-1)) - 1 (1\cdot1 - (-1)\cdot3) + 3 (1\cdot(-1) - 3\cdot3) = 2 (3 - 1) - 1 (1 - 3) + 3 (-1 - 9) = 2\cdot2 - 1\cdot(-2) + 3\cdot(-10) = 4 + 2 - 30 = -24 \] Since \( M_3 < 0 \), matrix A is not positive definite.

- Check if Matrix B is Symmetric

Check if matrix B is symmetric by comparing it to its transpose. Matrix B: \[ B = \begin{pmatrix} 5 & 0 & \sqrt{3} \ 0 & 3 & 0 \ \sqrt{3} & 0 & 3 \end{pmatrix} \] Its transpose: \[ B^T = \begin{pmatrix} 5 & 0 & \sqrt{3} \ 0 & 3 & 0 \ \sqrt{3} & 0 & 3 \end{pmatrix} \] Since \( B = B^T \), matrix B is symmetric.

- Compute the Leading Principal Minors of Matrix B

Compute the leading principal minors to check for positive definiteness. The leading principal minors should be positive: \( M_1 = 5 \), \( M_2 = \det \begin{pmatrix} 5 & 0 \ 0 & 3 \end{pmatrix} = 5 \cdot 3 - 0 \cdot 0 = 15 \), \( M_3 = \det \begin{pmatrix} 5 & 0 & \sqrt{3} \ 0 & 3 & 0 \ \sqrt{3} & 0 & 3 \end{pmatrix} \). For \( M_3 \): \[ \begin{vmatrix} 5 & 0 & \sqrt{3} \ 0 & 3 & 0 \ \sqrt{3} & 0 & 3 \end{vmatrix} = 5 \begin{vmatrix} 3 & 0 \ 0 & 3 \end{vmatrix} - 0 \begin{vmatrix} 0 & 0 \ \sqrt{3} & 3 \end{vmatrix} + \sqrt{3} \begin{vmatrix} 0 & 3 \ \sqrt{3} & 0 \end{vmatrix} = 5 (3 \cdot 3 - 0 \cdot 0) = 45 \] Since \( M_1 > 0 \), \( M_2 > 0 \), and \( M_3 > 0 \), matrix B is positive definite.

- Perform Cholesky Decomposition on Matrix B

Since matrix B is positive definite and symmetric, we can perform Cholesky decomposition to find the lower triangular matrix L such that \( B = LL^T \).\[ L = \begin{pmatrix} l_{11} & 0 & 0 \ l_{21} & l_{22} & 0 \ l_{31} & l_{32} & l_{33} \end{pmatrix} \] We solve for the elements of L by equating \( B = LL^T \): \[ LL^T = \begin{pmatrix} l_{11}^2 & l_{11} l_{21} & l_{11} l_{31} \ l_{11} l_{21} & l_{21}^2 + l_{22}^2 & l_{21} l_{31} + l_{22} l_{32} \ l_{11} l_{31} & l_{21} l_{31} + l_{22} l_{32} & l_{31}^2 + l_{32}^2 + l_{33}^2 \end{pmatrix} \]\( l_{11} = \sqrt{5} \), \( l_{21} = 0 \), \( l_{22} = \sqrt{3} \), \( l_{31} = \sqrt{3}/ \sqrt{5} \) and \( l_{32} = 0 \) and \( l_{33} = \sqrt{3 - 3/5} = \sqrt{12/5} \). Therefore, \[ L = \begin{pmatrix} \sqrt{5} & 0 & 0 \ 0 & \sqrt{3} & 0 \ \sqrt{3}/\sqrt{5} & 0 & \sqrt{12/5} \end{pmatrix} \]

Key concepts

Positive Definiteness

Understanding whether a matrix is positive definite is crucial in many areas of computations, especially for Cholesky decomposition. A matrix is considered positive definite if, for any non-zero vector \(\boldsymbol{x}\), the quadratic form \(\boldsymbol{x}^T \boldsymbol{A} \boldsymbol{x} > 0\).
Positive definite matrices have several important properties:

• All eigenvalues are positive.
• The matrix is symmetric.
• All leading principal minors are positive.

In summary, positive definiteness ensures that the matrix behaves nicely under various mathematical operations, making it suitable for Cholesky decomposition.

Symmetric Matrices

A symmetric matrix is one that is equal to its transpose, i.e., \( \boldsymbol{A} = \boldsymbol{A}^T \). This means
that the matrix is mirrored across its main diagonal.
For example, the matrix \(\begin{pmatrix} 2& 1& 3 \ 1& 3& -1 \ 3& -1& 1 \ \end{pmatrix}\) is symmetric.
Symmetric matrices have special properties:

• Eigenvalues are real.
• They can be decomposed using Cholesky decomposition if they are also positive definite.

Symmetric matrices are essential because many mathematical operations and optimizations
assume or require symmetry, making problems easier to solve.

Leading Principal Minors

Leading principal minors are the determinants of the upper-left submatrices. For instance,consider the matrix \(\boldsymbol{A}\): \(\begin{pmatrix} 2& 1& 3 \ 1& 3& -1 \3& -1& 1 \ \end{pmatrix}\).
Its leading principal minors are:

• \(M_1 = 2\)
• \(M_2 = \text{det}(\begin{pmatrix} 2& 1 \ 1& 3 \ \end{pmatrix}) = 5\)
• \(M_3 = \text{det}(\begin{pmatrix} 2& 1& 3 \ 1& 3& -1 \ 3& -1& 1 \ \end{pmatrix}) = -24\)

A matrix is positive definite if all its leading principal minors are positive. This is why wecompute these to check for positive definiteness.
In our example, since \(M_3 < 0\), matrix \(\boldsymbol{A}\) is not
positive definite.

Lower Triangular Matrix

In Cholesky decomposition, we decompose a positive definite, symmetric matrix \( \boldsymbol{A} \) into the product \(\boldsymbol{LL^T}\). Here, \( \boldsymbol{L} \) is a lower triangular matrix. A lower triangular matrix has all elements
above the main diagonal equal to zero. For instance,\(\begin{pmatrix} \ l_{11} & 0 & 0 \ \ l_{21} & l_{22} & 0 \ \ l_{31} & l_{32} & l_{33} \ \end{pmatrix} \). Lower triangular matrices have useful properties:

• Easy computation of determinants.
• Efficient for solving linear systems.

In our example, we found the lower triangular matrix for matrix
\( \boldsymbol{B} \):\( \begin{pmatrix} \ \sqrt{5} & 0 & 0 \ \0 & \sqrt{3} & 0 \ \frac{\sqrt{3}}{\sqrt{5}} & 0 & \sqrt{\frac{12}{5}} \ \end{pmatrix} \). This matrix is crucial for various computational purposes.