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Let \(A=L U\) be an LU factorization. Explain why \(A\) can be row reduced to \(U\) using only replacement operations. (This fact is the converse of what was proved in the text.)

Short Answer

Expert verified
A can be row reduced to U using only replacement operations, as matrix L in LU factorization encodes these operations.

Step by step solution

01

Understanding LU Factorization

In LU factorization, a matrix \(A\) is expressed as the product of a lower triangular matrix \(L\) and an upper triangular matrix \(U\). This means that \(A = LU\). Matrix \(L\) is a reflection of the row operations that reduce \(A\) to \(U\), specifically the replacement operations.
02

Properties of Matrix L

The matrix \(L\) is a lower triangular matrix with 1's on the diagonal and elements below the diagonal that correspond to the multipliers used during the row reduction process from \(A\) to \(U\). These are precisely the scalars used in the replacement operations.
03

Using L to Represent Row Operations

The elements in \(L\) allow us to reconstruct the row operations performed during Gaussian elimination: each element below the diagonal in \(L\) represents a specific replacement operation needed to transform \(A\) into \(U\). Thus, \(A\) can be reduced to \(U\) using these stored operations.
04

Verifying Row Reduction Operations

By expressing \(A\) as \(LU\), it is evident that one can sequentially apply the inverse of the operations stored in \(L\) to \(A\) to achieve \(U\) without applying any additional row swaps or scaling beyond initial setup. These involve only replacing rows, which validates the fact that \(A\) can be reduced to \(U\) using just the replacement operations.

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Key Concepts

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

Gaussian elimination
Gaussian elimination is a method used to solve systems of linear equations. The process involves performing row operations on a matrix to transform it into an upper triangular form or even further to row echelon form. During Gaussian elimination, we use a sequence of operations to simplify the matrix, making it easier to find solutions to the system of equations.
  • Steps of Gaussian Elimination:
    • Identify the leftmost column that is not completely zero.
    • Select a pivot (a non-zero element) and interchange rows, if necessary, to move this pivot to the top of the column.
    • Use row replacement operations to create zeros below the pivot in its column.
    • Repeat this process for each pivot in successive columns until the matrix is upper triangular or in row echelon form.
Remember, Gaussian elimination relies on row operations to systematically simplify the matrix and does not alter the solution set of the associated linear system. This method forms the foundation of LU factorization, where row operations are recorded to help decompose a matrix into its lower and upper triangular constituents.
lower triangular matrix
A lower triangular matrix is a type of square matrix where all the elements above the principal diagonal are zero. This means only elements on and below this diagonal can be non-zero. In LU factorization, the matrix L is lower triangular, recording the multipliers used during the elimination process.
  • Characterizing Lower Triangular Matrices:
    • Diagonal elements are typically ones in LU factorization.
    • All non-diagonal, non-zero elements are below the main diagonal.
    • These matrices are used to store the effects of row operations in Gaussian elimination.
The lower triangular matrix L illustrates the replacement operations necessary to reduce a given matrix A into its upper triangular form U. Hence, it effectively captures the essence of the pathway through which A transitions in the elimination process.
upper triangular matrix
An upper triangular matrix is a type of matrix in which all the elements below the principal diagonal are zero. This configuration simplifies solving systems of equations by back-substitution. In the context of LU factorization, the matrix U is the target form achieved after applying Gaussian elimination to matrix A.
  • Features of Upper Triangular Matrices:
    • Entries along the diagonal can vary and are generally non-zero.
    • Facilitates easy computation of determinants and solutions of linear equations.
    • Results from the process of eliminating variables from a matrix, bringing it to a simpler state.
Understanding the structure and role of the upper triangular matrix is essential, as it allows us to see the final "simplified" form of the matrix after row operations, making it easier to determine the solutions to linear systems.
row operations
Row operations are fundamental transformations used in the manipulation of matrices and are integral to the Gaussian elimination process. They allow us to transform a matrix while preserving the solution set of its system of equations.
  • Types of Row Operations:
    • Row Replacement: Adding or subtracting multiples of one row from another to introduce zeros and create simpler forms.
    • Row Interchange: Swapping two rows to position a better pivot element, though not used in LU factorization for row reduction to U.
    • Row Scaling: Multiplying all elements of a row by a non-zero constant.
In LU factorization, the focus is primarily on row replacement operations. The lower triangular matrix L records these specific operations, reflecting how the elimination process transformed matrix A into its upper triangular form U. These operations are reversed to verify the transformation without altering the fundamental solutions of the system.

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Most popular questions from this chapter

In Exercises 21 and \(22,\) mark each statement True or False. Justify each answer. a. A subspace of \(\mathbb{R}^{n}\) is any set \(H\) such that (i) the zero vector is in \(H,\) (ii) \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{u}+\mathbf{v}\) are in \(H,\) and (iii) \(c\) is a scalar and \(c \mathbf{u}\) is in \(H .\) b. If \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\) are in \(\mathbb{R}^{n},\) then \(\operatorname{Span}\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is the same as the column space of the matrix \(\left[\mathbf{v}_{1} \cdots \mathbf{v}_{p}\right] .\) c. The set of all solutions of a system of \(m\) homogeneous equations in \(n\) unknowns is a subspace of \(\mathbb{R}^{m} .\) d. The columns of an invertible \(n \times n\) matrix form a basis for \(\mathbb{R}^{n}\) . e. Row operations do not affect linear dependence relations among the columns of a matrix.

Let \(T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) be an invertible linear transformation, and let \(S\) and \(U\) be functions from \(\mathbb{R}^{n}\) into \(\mathbb{R}^{n}\) such that \(S(T(\mathbf{x}))=\mathbf{x}\) and \(U(T(\mathbf{x}))=\mathbf{x}\) for all \(\mathbf{x}\) in \(\mathbb{R}^{n} .\) Show that \(U(\mathbf{v})=S(\mathbf{v})\) for all \(\mathbf{v}\) in \(\mathbb{R}^{n} .\) This will show that \(T\) has a unique inverse, as asserted in Theorem \(9 .[\text { Hint: Given any }\) \(\mathbf{v}\) in \(\mathbb{R}^{n},\) we can write \(\mathbf{v}=T(\mathbf{x})\) for some \(\mathbf{x} .\) Why? Compute \(S(\mathbf{v})\) and \(U(\mathbf{v}) . ]\)

Let \(C\) be an \(n \times n\) consumption matrix whose column sums are less than \(1 .\) Let \(\mathbf{x}\) be the production vector that satisfies a final demand \(\mathbf{d},\) and let \(\Delta \mathbf{x}\) be a production vector that satisfies a different final demand \(\Delta \mathbf{d}\) . a. Show that if the final demand changes from d to d \(+\Delta \mathbf{d},\) then the new production level must be \(\mathbf{x}+\Delta \mathbf{x}\) . Thus \(\Delta \mathbf{x}\) gives the amounts by which production must change in order to accommodate the change \(\Delta \mathbf{d}\) in demand. b. Let \(\Delta \mathbf{d}\) be the vector in \(\mathbb{R}^{n}\) with 1 as the first entry and \(0^{\prime}\) s elsewhere. Explain why the corresponding production \(\Delta \mathbf{x}\) is the first column of \((I-C)^{-1} .\) This shows that the first column of \((I-C)^{-1}\) gives the amounts the various sectors must produce to satisfy an increase of 1 unit in the final demand for output from sector \(1 .\)

In Exercises 19–24, justify each answer or construction. Construct a \(4 \times 3\) matrix with rank 1

In Exercises \(3-6,\) the vector \(\mathbf{x}\) is in a subspace \(H\) with a basis \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\\} .\) Find the \(\mathcal{B}\) -coordinate vector of \(\mathbf{x} .\) $$ \mathbf{b}_{1}=\left[\begin{array}{r}{1} \\ {-3}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{-3} \\ {5}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{-7} \\ {5}\end{array}\right] $$

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