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Chapter 1: Problem 22

In the Cholesky factorization \(S=A^{\mathrm{T}} A\), with \(A=\sqrt{D} L^{\mathrm{T}}\), the square roots of the pivots are on the diagonal of \(A\). Find \(A\) (upper triangular) for $$ S=\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 8 \end{array}\right] \quad \text { and } \quad S=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 7 \end{array}\right] $$

Short Answer

Expert verified
For matrix 1, \( A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{bmatrix} \). For matrix 2, \( A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & \sqrt{5} \end{bmatrix} \).

Step by step solution

01

Check Symmetry and Positivity

Ensure each matrix \( S \) is symmetric and positive definite. Both matrices are symmetric by definition. We assume they are positive definite for the Cholesky factorization to be applicable.
02

Cholesky Factorization of First Matrix

For \( S = \begin{bmatrix} 9 & 0 & 0 \ 0 & 1 & 2 \ 0 & 2 & 8 \end{bmatrix} \), \( A \) should be an upper triangular matrix. Start by setting \( A = \begin{bmatrix} a & b & c \ 0 & d & e \ 0 & 0 & f \end{bmatrix} \). Compare with \( S = A^T A \), so \( a^2 = 9 \), \( d^2 + e^2 = 8 \), \( e \times f = 2 \). Solve these to get \( a = 3 \), \( d = 1 \), \( e = 2 \), and \( f = \sqrt{4} = 2 \). Thus, \( A = \begin{bmatrix} 3 & 0 & 0 \ 0 & 1 & 2 \ 0 & 0 & 2 \end{bmatrix} \).
03

Cholesky Factorization of Second Matrix

For \( S = \begin{bmatrix} 1 & 1 & 1 \ 1 & 2 & 2 \ 1 & 2 & 7 \end{bmatrix} \), set \( A = \begin{bmatrix} a & b & c \ 0 & d & e \ 0 & 0 & f \end{bmatrix} \). From \( A^T A = S \), solve for the elements as follows: \( a^2 = 1 \), \( a \times b = 1 \), \( a \times c + b \times e = 1 \), \( b^2 + d^2 = 2 \), \( b \times c + d \times e = 2 \), and \( e \times f + c^2 = 7 \). This gives \( a = 1 \), \( b = 1 \), \( c = 0 \), \( d = 1 \), \( e = 1 \), and \( f = \sqrt{5} \). Thus, \( A = \begin{bmatrix} 1 & 1 & 0\ 0 & 1 & 1 \ 0 & 0 & \sqrt{5} \end{bmatrix} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Positive Definite Matrices
Positive definite matrices are special types of matrices that have unique properties making them important in matrix decompositions like the Cholesky factorization. For a matrix to be considered positive definite, all its eigenvalues must be positive. This ensures that any non-zero vector \( x \) gives a strictly positive result when we compute \( x^T S x \).
  • These matrices are always symmetric, which means \( S = S^T \).
  • They are pivotal in ensuring the existence of the Cholesky decomposition, as only positive definite matrices can be decomposed this way.
  • In practical terms, positive definite matrices often represent systems where energy or variance is always increasing, never decreasing or stagnating.
Understanding whether a matrix is positive definite is crucial before beginning the process of Cholesky factorization. This ensures the factorization is properly defined and can be carried out smoothly.
Upper Triangular Matrix
An upper triangular matrix is a matrix where all the entries below the main diagonal are zero. This structure simplifies many algebraic operations. In the context of Cholesky factorization, the matrix \( A \) is expressed as an upper triangular matrix.
  • It has the form \( A = \begin{bmatrix} a & b & c \ 0 & d & e \ 0 & 0 & f \end{bmatrix} \).
  • The diagonal elements are crucial as they are the square roots of the pivots derived from the Cholesky process.
  • Upper triangular matrices make solving equations using back substitution more straightforward.
In any matrix decomposition process like Cholesky's, arriving at an upper triangular form helps in reducing complexity and extracting pertinent information efficiently.
Symmetric Matrix
Symmetric matrices are square matrices that are identical to their transpose. In other words, a matrix \( S \) is symmetric if \( S = S^T \). This symmetry is a fundamental property that enables Cholesky factorization.
  • Every entry \( s_{i,j} \) equals \( s_{j,i} \), which confines the relationship to half the matrix.
  • Symmetry implies that the matrix is often balanced in terms of its representations of changes across dimensions.
  • For Cholesky factorization to work, given matrix \( S \) must be symmetric and positive definite.
Recognizing the symmetry of a matrix can significantly reduce computational effort, making matrix manipulations both convenient and efficient.
Matrix Decomposition
Matrix decomposition involves breaking down a matrix into a product of simpler matrices. In the case of Cholesky factorization, a symmetric positive definite matrix \( S \) is decomposed into the product of a lower triangular matrix and its transpose. This kind of decomposition reveals important structural insights about the matrix.
  • Cholesky decomposition specifically breaks a matrix \( S = A^T A \), allowing for efficient solutions in numerical problems.
  • Decomposition helps in simplifying complex systems and solving multiple equations with shared coefficients.
  • It has applications in optimization problems, simulations, and numerical methods where understanding the matrix's inner structure is beneficial.
Grasping the principles of matrix decomposition ensures a solid foundation for tackling more advanced topics in linear algebra and computational mathematics.

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Most popular questions from this chapter

Suppose \(S^{\mathrm{T}}=S\) and \(S x=\lambda x\) and \(S y=\alpha y\) are all real. Show that \(y^{\mathrm{T}} S \boldsymbol{x}=\lambda \boldsymbol{y}^{\mathrm{T}} \boldsymbol{x} \quad\) and \(\quad \boldsymbol{x}^{\mathrm{T}} S \boldsymbol{y}=\alpha \boldsymbol{x}^{\mathrm{T}} \boldsymbol{y} \quad\) and \(\quad \boldsymbol{y}^{\mathrm{T}} S \boldsymbol{x}=\boldsymbol{x}^{\mathrm{T}} S \boldsymbol{y} .\) Show that \(y^{\mathrm{T}} x\) must be zero if \(\lambda \neq \alpha:\)

Start with a matrix \(B\). If we want to take combinations of its rows, we premultiply by \(A\) to get \(A B\). If we want to take combinations of its columns, we postmultiply by \(C\) to get \(B C\). For this question we will do both. Row operations then column operations First \(A B\) then \((A B) C\) Column operations then row operations First \(B C\) then \(A(B C)\) The associative law says that we get the same final result both ways. Verify \((A B) C=A(B C)\) for \(A=\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right] \quad B=\left[\begin{array}{ll}b_{1} & b_{2} \\ b_{3} & b_{4}\end{array}\right] \quad C=\left[\begin{array}{ll}1 & 0 \\ c & 1\end{array}\right]\).

Find the SVD of the rank 1 matrix \(A=\left[\begin{array}{ll}2 & 4 \\ 1 & 2\end{array}\right]\). Factor \(A^{\mathrm{T}} A\) into \(Q \Lambda Q^{\mathrm{T}}\).

Suppose \(A\) and \(B\) have the same eigenvalues \(\lambda_{1, \ldots, \lambda_{n}}\) with the same independent eigenvectors \(x_{1}, \ldots, x_{n}\). Then \(A=B\). Reasan: Any vector \(x\) is a combination \(c_{1} x_{1}+\cdots+c_{n} x_{n}\). What is \(A x\) ? What is \(B x\) ?

Which of \(S_{1}, S_{2}, S_{3}, S_{4}\) has two positive eigenvalues? Use a test, don't compute the \(\lambda\) 's. Also find an \(x\) so that \(x^{\mathrm{T}} S_{1} x<0\), so \(S_{1}\) is not positive definite. $$ S_{1}=\left[\begin{array}{ll} 5 & 6 \\ 6 & 7 \end{array}\right] \quad S_{2}=\left[\begin{array}{ll} -1 & -2 \\ -2 & -5 \end{array}\right] \quad S_{3}=\left[\begin{array}{rr} 1 & 10 \\ 10 & 100 \end{array}\right] \quad S_{4}=\left[\begin{array}{rr} 1 & 10 \\ 10 & 101 \end{array}\right] \text {. } $$
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