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Chapter 8: Problem 10

Using the Choleski method, calculate the decomposition \(A=L L^{T}\) for (a) \(\left[\begin{array}{rrr}2.25 & -3.0 & 4.5 \\ -3.0 & 5.0 & -10.0 \\ 4.5 & -10.0 & 34.0\end{array}\right]\) (b) \(\left[\begin{array}{rrrr}15 & -18 & 15 & -3 \\ -18 & 24 & -18 & 4 \\ 15 & -18 & 18 & -3 \\ -3 & 4 & -3 & 1\end{array}\right]\)

Short Answer

Expert verified
For part (a), matrix L is \( \left[ \begin{array}{ccc} 1.5 & 0 & 0 \ -2.0 & 1.0 & 0 \ 3.0 & -4.0 & 3.0 \end{array} \right] \). For part (b), decomposition is not fully possible.

Step by step solution

01

Define the matrices

Let matrix A be the one we want to decompose. For part (a), \( A = \left[\begin{array}{rrr}2.25 & -3.0 & 4.5 \ -3.0 & 5.0 & -10.0 \ 4.5 & -10.0 & 34.0\end{array}\right] \)
02

Initialize matrix L

We will initialize a lower triangular matrix \( L \) such that \( A = L L^T \). For a 3x3 matrix, \( L = \left[\begin{array}{ccc}l_{11} & 0 & 0 \ l_{21} & l_{22} & 0 \ l_{31} & l_{32} & l_{33} \end{array}\right] \)
03

Calculate L elements (part a - l11, l21, l31)

Calculate the elements of \( L \) by using the equations for Choleski decomposition: \( l_{11} = \sqrt{A_{11}} = \sqrt{2.25} = 1.5 \)\( l_{21} = \frac{A_{21}}{l_{11}} = \frac{-3.0}{1.5} = -2.0 \)\( l_{31} = \frac{A_{31}}{l_{11}} = \frac{4.5}{1.5} = 3.0 \)
04

Calculate L elements (part a - l22, l32)

Continue calculating: \( l_{22} = \sqrt{A_{22} - l_{21}^2} = \sqrt{5.0 - (-2.0)^2} = \sqrt{1.0} = 1.0 \)\( l_{32} = \frac{A_{32} - l_{31}l_{21}}{l_{22}} = \frac{-10.0 - (3.0)(-2.0)}{1.0} = -4.0 \)
05

Calculate L elements (part a - l33)

Finish calculating: \( l_{33} = \sqrt{A_{33} - l_{31}^2 - l_{32}^2} = \sqrt{34.0 - 3.0^2 - (-4.0)^2} = \sqrt{9.0} = 3.0 \)Therefore, \( L = \left[\begin{array}{ccc} 1.5 & 0 & 0 \ -2.0 & 1.0 & 0 \ 3.0 & -4.0 & 3.0 \end{array}\right] \)
06

Define and Initialize matrix L for part (b)

For part (b), matrix A is \( A = \left[\begin{array}{rrrr}15 & -18 & 15 & -3 \ -18 & 24 & -18 & 4 \ 15 & -18 & 18 & -3 \ -3 & 4 & -3 & 1\end{array}\right] \). Initialize \( L \) as a 4x4 lower triangular matrix: \( L = \left[\begin{array}{cccc}l_{11} & 0 & 0 & 0 \ l_{21} & l_{22} & 0 & 0 \ l_{31} & l_{32} & l_{33} & 0 \ l_{41} & l_{42} & l_{43} & l_{44} \end{array}\right] \)
07

Calculate L elements (part b - l11, l21, l31, l41)

Calculate the first column of L:\( l_{11} = \sqrt{A_{11}} = \sqrt{15} \approx 3.872 \)\( l_{21} = \frac{A_{21}}{l_{11}} = \frac{-18}{3.872} \approx -4.65 \)\( l_{31} = \frac{A_{31}}{l_{11}} = \frac{15}{3.872} \approx 3.872 \)\( l_{41} = \frac{A_{41}}{l_{11}} = \frac{-3}{3.872} \approx -0.774 \)
08

Calculate L elements (part b - l22, l32, l42)

Calculate the second column of L:\( l_{22} = \sqrt{A_{22} - l_{21}^2} = \sqrt{24 - (-4.65)^2} \approx 2 \)\( l_{32} = \frac{A_{32} - l_{31}l_{21}}{l_{22}} = \frac{-18 - 3.872(-4.65)}{2} \approx 0 \)\( l_{42} = \frac{A_{42} - l_{41}l_{21}}{l_{22}} = \frac{4 - (-0.774)(-4.65)}{2} \approx 1 \)
09

Calculate L elements (part b - l33, l43)

Continue calculating the third column of L:\( l_{33} = \sqrt{A_{33} - l_{31}^2 - l_{32}^2} = \sqrt{18 - 3.872^2 - 0^2} \approx 0 \)\( l_{43} = \frac{A_{43} - l_{41}l_{31} - l_{42}l_{32}}{l_{33}} = \frac{-3 - (-0.774)(3.872) - 1(0)}{0} \rightarrow \text{undefined, so we stop here} \)
10

Combine results

For part (a), \( L = \left[\begin{array}{ccc} 1.5 & 0 & 0 \ -2.0 & 1.0 & 0 \ 3.0 & -4.0 & 3.0 \end{array}\right] \)For part (b), decomposition is not possible past the first few rows because the second and third elements in the third row and column nullify further calculations.

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Key Concepts

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

matrix decomposition
Matrix decomposition is a powerful concept in linear algebra. It involves breaking down a matrix into simpler, easier-to-manage components. This can make complex matrix operations more efficient. For Choleski decomposition, we decompose a symmetric, positive-definite matrix into the product of a lower triangular matrix and its transpose.
This is written as:
  • Given matrix A, find matrices L such that:
    • A = LLT
Matrix decomposition plays a prominent role in solving systems of linear equations, improving computational efficiency in numerical studies, and stabilizing algorithms. By separating a matrix into its constituent parts, complex problems become more manageable.
linear algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It is fundamental to modern scientific and engineering analysis.
Essential concepts include vectors, matrices, determinants, eigenvalues, and eigenvectors, which are crucial for understanding advanced topics such as machine learning, quantum computing, and data analysis.
In matrix decomposition, we often transform complex matrix equations into simpler, solvable forms. For instance, Choleski decomposition, which we are discussing here, breaks down a matrix to facilitate easier matrix manipulations. Understanding the properties of matrices, such as symmetry and positive definiteness, is vital in linear algebra to apply the correct techniques.
numerical analysis
Numerical analysis is the study of algorithms that use numerical approximations for mathematical problems. It is important for implementing practical and efficient algorithms for computational tasks.
Key aspects include error analysis, convergence rates, and stability of numerical methods. When performing matrix decomposition, these factors ensure that the results are accurate and usable.
Methods like Choleski decomposition help reduce computational complexity, providing more efficient solutions for large matrix operations commonly encountered in scientific computing, engineering, and applied mathematics.
Evaluating the computational complexity and efficiency of an algorithm, such as Choleski decomposition, often requires understanding concepts from numerical analysis. Ensuring algorithms are optimal and error-resistant is a major goal.
Choleski method step-by-step
The Choleski method is a specific type of matrix decomposition used for symmetric, positive-definite matrices. Let's break down the steps:
1. **Initialize Matrix L:** Assume a lower triangular matrix L.
2. **Calculate Elements of L:** Use the given formulas to compute each element.
For example:
  • For matrix A in part (a):

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    • Calculate l11:

      • l11 = 鈭欰11 = 鈭�2.25 = 1.5
    • Calculate l21:
      • l21 = A21 / l11 = -3.0 / 1.5 = -2.0
    • Calculate l31:
      • l31 = A31 / l11 = 4.5 / 1.5 = 3.0
    • Continue this process to fill in L using the Choleski formulas.
Each step ensures the resulting lower triangular matrix L, when multiplied by its transpose LT, reconstructs matrix A. This method simplifies many linear algebra problems, making them solvable with efficient computational techniques.

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Most popular questions from this chapter

There are families of linear systems \(A_{k} x=b\) in which \(A_{k}\) changes in some simple way into a matrix \(A_{k+1}\), and it may then be simpler to find the \(L U\) factorization of \(A_{k+1}\) by modifying that of \(A_{k} .\) As an example that arises in the simplex method for linear programming, let \(A_{1}=\left[a_{1}, \ldots, a_{n}\right]\) and \(A_{2}=\left[a_{2}, \ldots, a_{n+1}\right]\), with all \(a_{j} \in \mathbf{R}^{n}\). Suppose \(A_{1}=L_{1} U_{1}\) is known, with \(L_{1}\) lower triangular and \(U_{1}\) upper triangular. Find a simple way to obtain the \(L U\) factorization \(A_{2}=L_{2} U_{2}\) from that for \(A_{1}\), assuming pivoting is not needed. Hint: Using \(L_{1} u_{i}=a_{i}, 1 \leq i \leq n\), write $$ A_{2}=L_{1}\left[u_{2}, u_{3}, \ldots u_{n}, L_{1}^{-1} a_{n+1}\right] \equiv L_{1} \tilde{U} $$ Show that \(\tilde{U}\) can be simply modified into an upper triangular form \(U_{2}\), and that this corresponds to the conversion of \(L_{1}\) into the desired \(L_{2}\). More precisely, \(U_{2}=M \tilde{U}, L_{2}=L_{1} M^{-1}\). Show that the operation cost for obtaining \(A_{2}=L_{2} U_{2}\) is \(O\left(n^{2}\right)\).

Let \(A\) be nonsingular. Let. \(A=L U=L D M\), with all \(l_{i i}, m_{i i}=1\), and \(D\) diagonal. Further assume \(A\) is symmetric. Show that \(M=L^{T}\), and thus \(A=L D L^{T} .\) Show \(A\) is positive definite if and only if all \(d_{i i}>0 .\)

Consider solving the integral equation $$ \lambda x(s)-\int_{0}^{1} \cos (\pi s t) x(t) d t=1 \quad 0 \leq s \leq 1 $$ by discretizing the integral with the midpoint numerical integration rule (5.2.18). More precisely, let \(n>0, h=1 / n, t_{i}=\left(i-\frac{1}{2}\right) h\) for \(i=1, \ldots, n\). We solve for approximate values of \(x\left(t_{1}\right), \ldots, x\left(t_{n}\right)\) by solving the linear system $$ \lambda z_{i}-\sum_{j=1}^{n} h \cos \left(\pi t_{i} t_{j}\right) z_{j}=1 \quad i=1, \ldots, n $$ Denote this linear system by \(\left(\lambda I-K_{n}\right) z=b\), with \(K_{n}\) of order \(n \times n\), $$ \left(K_{n}\right)_{i j}=h \cdot \cos \left(\pi t_{i} t_{j}\right) \quad b_{i}=1 \quad 1 \leq i, j \leq n $$ For sufficiently large \(n, z_{i} \doteq x\left(t_{i}\right), 1 \leq i \leq n .\) The value of \(\lambda\) is nonzero, and it is assumed here to not be an eigenvalue of \(K_{n}\). Solve \(\left(\lambda I-K_{n}\right) z=b\) for several values of \(n\), say \(n=2,4,8,16,32,64\) and print the vector solutions \(z\). If possible, also graph these solutions, to gain some idea of the solution function \(x(s)\) of the original integral equation. Use \(\lambda=4,2,1, .5\)

Let \(A\) be symmetric, positive definite, and of order \(n\). Let \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) be an \(A\) -orthogonal set in \(\mathbf{R}^{n}\), with all \(v_{i} \neq 0\). Define $$ Q_{j}=\frac{v_{j} v_{j}^{T} A}{v_{j}^{T} A v_{j}} \quad j=1, \ldots, n $$ Showing the following properties for \(Q_{j}\) 1\. \(Q_{j} v_{i}=0\) if \(i \neq j ;\) and \(Q_{j} v_{j}=v_{j}\) 2\. \(Q_{j}^{2}=Q_{j}\) 3\. \(\left(x, Q_{j} y\right)_{A}=\left(Q_{j} x, y\right)_{A}\), for all \(x, y \in \mathbf{R}^{n} .\) 4\. \(\left(Q_{j} x,\left(I-Q_{j}\right) y\right)_{A}=0\), for all \(x, y \in \mathbf{R}^{n}\) 5\. \(\|x\|_{A}^{2}=\left\|Q_{j} x\right\|_{A}^{2}+\left\|\left(I-Q_{j}\right) x\right\|_{A}^{2}\), for all \(x \in \mathbf{R}^{n}\). Properties (2)-(5) say that \(Q_{j}\) is an orthogonal projection on the vector space \(\mathbf{R}^{n}\). with the inner product \((\cdot, \cdot)_{A}\). Define $$ S_{k}=\operatorname{Span}\left\\{v_{1}, \ldots, v_{k}\right\\} $$ Show that the solution to the minimization problem $$ \underset{y \in S_{k}}{\operatorname{Min}}\|x-y\|_{A} $$ is given by $$ y=\left[\sum_{j=1}^{k} Q_{j}\right] x \equiv P_{k} x $$ The matrix \(P_{k}\) also satisfies properties (2)-(5).

Let \(A\) be real, symmetric, positive definite, and of order \(n\). Consider solving \(A x=b\) using Gaussian elimination without pivoting. The purpose of this problem is to justify that the pivots will be nonzero. (a) Show that all of the diagonal elements satisfy \(a_{i i}>0\). This shows that \(a_{11}\) can be used as a pivot element. (b) After elimination of \(x_{1}\) from equations 2 through \(n\), let the resulting matrix \(A^{(2)}\) be written as $$ A^{(2)}=\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ 0 & & & \\ \vdots & & \hat{A}^{(2)} & \\ 0 & & & \end{array}\right] $$ Show that \(\hat{A}^{(2)}\). is symmetric and positive definite. This procedure can be continued inductively to each stage of the elimination process, thus justifying the existence of nonzero pivots at ev茅ry step. Hint: To prove \(\hat{A}^{(2)}\) is positive definite, first prove the identity $$ \sum_{i, j=2}^{n} a_{i j}^{(2)} x_{i} x_{j}=\sum_{i, j=1}^{n} a_{i j} x_{i} x_{j}-a_{11}\left[x_{1}+\sum_{j=2}^{n} \frac{a_{j 1}}{a_{11}} x_{j}\right]^{2} $$ for any choice of \(x_{1}, x_{2}, \ldots, x_{n} .\) Then choose \(x_{1}\) suitably.
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