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Find an LU factorization of the matrices in Exercises \(7-16\) (with \(L\) unit lower triangular). Note that MATLAB will usually produce a permuted LU factorization because it uses partial pivoting for numerical accuracy. $$ \left[\begin{array}{ll}{6} & {9} \\ {4} & {5}\end{array}\right] $$

Short Answer

Expert verified
The LU factorization is \(L = \begin{pmatrix} 1 & 0 \\ \frac{2}{3} & 1 \end{pmatrix}\) and \(U = \begin{pmatrix} 6 & 9 \\ 0 & -1 \end{pmatrix}\).

Step by step solution

01

Understand LU Factorization

LU factorization of a matrix is a decomposition into a lower triangular matrix \(L\) and an upper triangular matrix \(U\), such that \(A = LU\). For this exercise, \(L\) will be a unit lower triangular matrix, meaning all diagonal elements are 1. The matrix provided is \( \begin{pmatrix} 6 & 9 \ 4 & 5 \end{pmatrix} \).
02

Set Up Matrices L and U

Start by initializing matrices \(L\) and \(U\). Matrix \(U\) will initially be the same as matrix \(A\), and \(L\) will be an identity matrix of the same size: \(L = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\). \[ U = \begin{pmatrix} 6 & 9 \ 4 & 5 \end{pmatrix} \]
03

Eliminate Elements Below Diagonal in U

To make \(U\) upper triangular, eliminate the element below the diagonal in the first column of \(U\). This means the second row, first column element of \(U\) should become zero. Divide the first row by \(6\) and subtract from the second row after scaling by \(\frac{4}{6}\):\[ \frac{4}{6} \times \begin{pmatrix} 6 & 9 \end{pmatrix} = \begin{pmatrix} 4 & 6 \end{pmatrix}\]Subtracting from second row:\[ \begin{pmatrix} 4 & 5 \end{pmatrix} - \begin{pmatrix} 4 & 6 \end{pmatrix} = \begin{pmatrix} 0 & -1 \end{pmatrix}\]Thus, \( U = \begin{pmatrix} 6 & 9 \ 0 & -1 \end{pmatrix} \).
04

Update L with Multipliers

Record the multiplier used for the row operation in \(L\). For \(L[2,1]\), it is \(\frac{4}{6} = \frac{2}{3}\). Update \(L\):\[ L = \begin{pmatrix} 1 & 0 \ \frac{2}{3} & 1 \end{pmatrix} \]
05

Verify A = LU

Multiply \(L\) and \(U\) to ensure \(A = LU\):\[ L = \begin{pmatrix} 1 & 0 \ \frac{2}{3} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} 6 & 9 \ 0 & -1 \end{pmatrix} \]Calculate the product:\[ \begin{pmatrix} 1 & 0 \ \frac{2}{3} & 1 \end{pmatrix} \begin{pmatrix} 6 & 9 \ 0 & -1 \end{pmatrix} = \begin{pmatrix} 6 & 9 \ \frac{2}{3} \times 6 + 0 & \frac{2}{3} \times 9 - 1 \end{pmatrix} = \begin{pmatrix} 6 & 9 \ 4 & 5 \end{pmatrix} \] This matches the original matrix \(A\). Thus, factors are correct.

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Key Concepts

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

Linear Algebra
Linear algebra is a branch of mathematics focusing on vector spaces and linear mappings between them. It deals extensively with vectors, matrices, and linear functions. Linear algebra is essential because it forms the foundation for topics such as matrix decomposition, which is key in solving many computational problems.

Matrices are a core component of linear algebra and serve multiple purposes, including representing systems of linear equations, transformations, and binary relations. Understanding these basic constructs helps in grasping more complex operations like the LU factorization.
  • Vectors and matrices: These are the fundamental elements of linear algebra.
  • Transformations: Using matrices to transform vectors in space.
  • Systems of equations: Matrices can represent and solve these systems.
Mastering linear algebra concepts makes it easier to explore and apply numerical solutions in scientific and engineering contexts.
Matrix Decomposition
Matrix decomposition involves breaking down a matrix into simpler, constituent components. In the context of linear algebra, this is crucial because it simplifies tasks such as solving matrix equations, computing determinants, and inverting matrices.

One of the most popular methods of matrix decomposition is LU factorization. LU factorization decomposes a square matrix into a product of a lower triangular matrix and an upper triangular matrix. This is highly useful for numerical analysis and computational algorithms, as it allows for a more efficient handling of linear equations.
  • LU Factorization: Decomposes a matrix into a lower and an upper triangular matrix, simplifying calculations.
  • Cholesky Decomposition: Used specifically for symmetric, positive definite matrices.
  • QR Decomposition: Breaks down a matrix into an orthogonal matrix and an upper triangular matrix.
Understanding these decomposition methods is pivotal in modern data science, machine learning, and computer graphics, where large matrices are frequently manipulated.
Upper Triangular Matrix
An upper triangular matrix is a type of square matrix where all the entries below the main diagonal are zero. This form is particularly useful in simplifying complex calculations, such as matrix multiplication and inverse calculation.

In an LU factorization, the matrix referred to as matrix "U" is the upper triangular matrix. It retains the right-hand part of the original matrix while the lower part becomes zero during the factorization process. Therefore, the element manipulation ensures all entries below the diagonal turn to zeros without altering the solution set.
  • Properties: Retains zeros below the diagonal, making computational tasks like solving equations more straightforward.
  • Applications: Facilitates easy back substitution in linear equations.
Utilizing upper triangular matrices efficiently requires understanding both their structure and how they interact with other matrices in system equations.
Lower Triangular Matrix
A lower triangular matrix is a square matrix in which all the entries above the main diagonal are zero. They play a crucial role in the LU decomposition process, representing the matrix "L" in the factorization.

When performing LU factorization, the lower triangular matrix "L" often takes the form of a unit lower triangular matrix, meaning it has ones on its main diagonal. This simplifies calculations and ensures stability in numerical solutions by maintaining the integrity of the original matrix structure.
  • Structure: Zeros above the diagonal, simplifying forward substitution algorithms.
  • Usage: Facilitates solving of matrix equations by reducing complexity.
Comprehending lower triangular matrices and their properties is crucial in fields requiring extensive use of numerical linear algebra, including engineering and physics.

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Most popular questions from this chapter

In Exercises \(3-8,\) find the \(3 \times 3\) matrices that produce the described composite 2 \(\mathrm{D}\) transformations, using homogeneous coordinates. Translate by \((3,1),\) and then rotate \(45^{\circ}\) about the origin.

In Exercises 17 and \(18,\) mark each statement True or False. Justify each answer. Here \(A\) is an \(m \times n\) matrix. a. If \(\mathcal{B}\) is a basis for a subspace \(H,\) then each vector in \(H\) can be written in only one way as a linear combination of the vectors in \(\mathcal{B}\) . bectors in \(\mathcal{B}\) . b. If \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is a basis for a subspace \(H\) of \(\mathbb{R}^{n},\) then the correspondence \(\mathbf{x} \mapsto[\mathbf{x}]_{\mathcal{B}}\) makes \(H\) look and act the same as \(\mathbb{R}^{p}\) . c. The dimension of Nul \(A\) is the number of variables in the equation \(A \mathbf{x}=\mathbf{0} .\) d. The dimension of the column space of \(A\) is rank \(A .\) e. If \(H\) is a \(p\) -dimensional subspace of \(\mathbb{R}^{n},\) then a linearly independent set of \(p\) vectors in \(H\) is a basis for \(H .\)

In Exercises 13 and 14, find a basis for the subspace spanned by the given vectors. What is the dimension of the subspace? $$ \left[\begin{array}{r}{1} \\ {-1} \\ {-2} \\\ {5}\end{array}\right],\left[\begin{array}{r}{2} \\ {-3} \\ {-1} \\\ {6}\end{array}\right],\left[\begin{array}{r}{0} \\ {2} \\ {-6} \\\ {8}\end{array}\right],\left[\begin{array}{r}{-1} \\ {4} \\ {-7} \\\ {7}\end{array}\right],\left[\begin{array}{r}{3} \\ {-8} \\ {9} \\\ {-5}\end{array}\right] $$

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: A \((k+1) \times(k+1)\) matrix \(A_{1}\) can be written in the form below, where \(a\) is a scalar, \(\mathbf{v}\) is in \(\mathbb{R}^{k},\) and \(A\) is a \(k \times k\) lower triangular matrix. See the Study Guide for help with induction. \(A_{1}=\left[\begin{array}{cc}{a} & {\mathbf{0}^{T}} \\ {\mathbf{v}} & {A}\end{array}\right]\)

In Exercises 37 and 38, construct bases for the column space and the null space of the given matrix A. Justify your work. $$ A=\left[\begin{array}{rrrrr}{5} & {2} & {0} & {-8} & {-8} \\ {4} & {1} & {2} & {-8} & {-9} \\ {5} & {1} & {3} & {5} & {19} \\ {-8} & {-5} & {6} & {8} & {5}\end{array}\right] $$

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