Problem 31 - Differential Equations and Linear Algebra

Question

QR Factorization: It can be shown that any invertible \(n \times n\) matrix has a factorization of the form $$A=Q R,$$ where \(Q\) and \(R\) are invertible, \(R\) is upper triangular, and \(Q\) satisfies \(Q^{T} Q=I_{n}\) (i.e., \(Q\) is orthogonal). Determine an algorithm for solving the linear system \(A \mathbf{x}=\mathbf{b}\) using this QR factorization.

Step-by-Step Solution

Answer

To solve the linear system \(A\textbf{x}=\textbf{b}\) using the QR factorization, follow these steps: 1. Compute the QR factorization of the given matrix \(A\) (using methods like Gram-Schmidt, Householder, or Givens rotations), resulting in \(A = QR\). 2. Rewrite the linear system using the factorization: \(QR\textbf{x} = \textbf{b}\). 3. Utilize the orthogonal property of \(Q\) to simplify the equation by multiplying both sides by \(Q^T\): \(R\textbf{x} = Q^T\textbf{b}\). 4. Solve for \(\textbf{x}\) using back-substitution, starting from the bottom row of the upper triangular matrix \(R\) and working upwards. The resulting \(\textbf{x}\) is the solution to the linear system \(A\textbf{x}=\textbf{b}\) using QR factorization.

Step 1: QR Factorization of A

To start, we need to find the factorization of the form \(A = QR\) where \(A\) is an invertible \(n \times n\) matrix, \(Q\) is an orthogonal matrix (i.e., \(Q^TQ = I_n\)), and \(R\) is an upper triangular matrix. There are various algorithms to compute the QR factorization, such as Gram-Schmidt, Householder, or Givens rotations. Choose one of these methods and compute the QR factorization of the given matrix \(A\).

Step 2: Rewrite the Linear System

Once we obtain the QR factorization of \(A\), we can rewrite the linear system \(A\textbf{x} = \textbf{b}\) using the factorization, resulting in the equation: \(Q R \textbf{x} = \textbf{b}\)

Step 3: Utilize the Orthogonal Property of Q

Since \(Q\) is orthogonal, \(Q^TQ = I_n\). Thus, multiply both sides of the equation \(QR\textbf{x} = \textbf{b}\) by \(Q^T\) from the left, \(Q^T(Q R \textbf{x}) = Q^T\textbf{b}\) This simplifies to: \(R\textbf{x} = Q^T\textbf{b}\)

Step 4: Solve for x using Back-substitution

As \(R\) is an upper triangular matrix, we can solve for \(\textbf{x}\) using the back-substitution method. Start from the bottom row of the matrix \(R\) and work your way upwards, substituting the already retrieved values of \(x_i\) back into the system until you have all of the components of \(\textbf{x}\). After the back-substitution, the resulting \(\textbf{x}\) will be the solution to the linear system \(A\textbf{x} = \textbf{b}\) using the QR factorization.

Key Concepts

Linear System

A linear system refers to a collection of linear equations involving the same set of variables. These systems can be represented in matrix form as \[ A\mathbf{x} = \mathbf{b} \] where \( A \)is the matrix of coefficients, \( \mathbf{x} \)is the vector of variables, and \( \mathbf{b} \)is the vector of constants.
The goal when dealing with a linear system is to determine the values in \( \mathbf{x} \)that satisfy all equations simultaneously.

Solving linear systems is a fundamental task in mathematics and its applications. By transforming the system using QR factorization, the solution process becomes more structured and organized. QR factorization involves decomposing matrix \( A \)into two matrices, \( Q \)and \( R \), making the subsequent calculations more convenient.

Orthogonal Matrix

An orthogonal matrix is a square matrix \( Q \)that satisfies the condition \( Q^TQ = I_n \), where \( Q^T \)is the transpose of \( Q \)and \( I_n \)is the identity matrix of the same size.
This property implies that the columns of an orthogonal matrix form an orthonormal basis.

Orthogonal matrices have several interesting properties:

• They preserve the length of vectors upon transformation. Thus, multiplying a vector by an orthogonal matrix doesn't change its magnitude.
• The determinant of an orthogonal matrix is always \( +1 \)or \( -1 \).
• They are inherently stable for numerical computations, reducing round-off errors significantly.

In the context of QR factorization, using an orthogonal matrix simplifies the problem since the inverse of an orthogonal matrix is simply its transpose. This makes it straightforward to manipulate the resulting equations when solving linear systems.

Back-substitution

Once we've rewritten our linear equation to \( R\textbf{x} = Q^T\textbf{b} \)and know that \( R \)is an upper triangular matrix, back-substitution is the method used to solve for \( \textbf{x} \).
Back-substitution is a technique where you start solving from the bottom of the matrix and move upwards.

Here’s a step-by-step guide on how it works:

• Identify the last row of the matrix \( R \), which will have the form \( r_{nn}x_n = d_n \), where \( x_n \)is the unknown variable, and \( d_n \)is the known term.
• Solve this equation for \( x_n \).
• Proceed to the second-to-last row, which has already \( x_n \)known, and solve for \( x_{n-1} \), and so on, until you reach the first row.
• Each row reduces the equation further until all variables are determined.

This method is efficient and particularly useful for solving systems derived from QR factorization, where \( R \)is an upper triangular matrix.

Upper Triangular Matrix

An upper triangular matrix is a type of square matrix where all the elements below the main diagonal are zero. This means any element \( a_{ij} \)for \( i > j \)is zero.
In mathematical terms: \[ R = \begin{pmatrix} r_{11} & r_{12} & \cdots & r_{1n} \ 0 & r_{22} & \cdots & r_{2n} \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & r_{nn} \end{pmatrix} \]

The benefits of using an upper triangular matrix such as \( R \)when solving linear systems include:

• Greatly simplifying the solution process, as shown in the back-substitution method.
• Making the algorithm computationally efficient because each step directly reduces the size of the problem.
• They are easy to work with since the determinant is simply the product of its diagonal elements.

In QR factorization, obtaining matrix \( R \)as upper triangular is crucial because it allows the subsequent steps of solving the system to focus only on the variables needed, simplifying calculations.