Solving Linear Equations: A Comprehensive Guide to QR Factorization

Solving systems of linear equations is a fundamental task in various fields such as engineering, physics, and computer science. One such sophisticated method involves QR factorization, a powerful numerical technique that can effectively tackle such problems. In this article, we will explore the steps and processes involved in solving linear equations using QR factorization, along with a practical example to illustrate the procedure.

Understanding the Problem

Consider a system of linear equations in the form:

[ A x b ]

where:

[ A ] is an [ m times n ] matrix with [ m geq n ], [ b ] is an [ m times 1 ] column vector, [ x ] is an [ n times 1 ] column vector of unknowns.

This system can be quite complex, especially when dealing with large matrices. However, employing QR factorization can simplify the solution process.

Performing QR Factorization

QR factorization decomposes the matrix [ A ] into the product of an orthogonal matrix [ Q ] and an upper triangular matrix [ R ]. This can be mathematically represented as:

[ A QR ]

In this context:

[ Q^T Q I ], the identity matrix, [ R ] is an upper triangular matrix.

There are several methods to perform QR factorization, including the Gram-Schmidt process, Householder transformations, or Givens rotations. Modern software libraries like NumPy in Python offer built-in functions to perform QR factorization efficiently.

Substituting [ A ] in the Equation

After performing QR factorization, substitute [ A ] in the original equation [ A x b ]:

[ Q R x b ]

This substitution can help transform the system into a more manageable form.

Solving for [ y ]

Let [ y R x ]. Then the equation becomes:

[ Q^T y Q^T b ]

Since [ Q ] is orthogonal, computing [ Q^T b ] is straightforward. This step simplifies our equation further.

Solving the Upper Triangular System

Now, solve the upper triangular system:

[ R y Q^T b ]

This can be done efficiently using back substitution, a common method for solving triangular matrices. Back substitution involves solving for the unknowns starting from the bottom of the matrix and moving upwards.

Solving for [ x ]

Finally, solve for [ x ] using:

[ x R^{-1} y ]

Due to the upper triangular nature of [ R ], this step can also be performed using back substitution.

Example

Consider the following system:

[ begin{align} 2x_1 x_2 1 4x_1 2x_2 2 end{align} ]

In matrix form:

[ A begin{bmatrix} 2 1 4 2 end{bmatrix} quad b begin{bmatrix} 1 2 end{bmatrix} ]

Perform QR factorization (assuming we get the following results):

[ Q begin{bmatrix} 0.4472 0.8944 0.8944 -0.4472 end{bmatrix} quad R begin{bmatrix} 4.4721 2.2361 0 0 end{bmatrix} ]

Compute [ Q^T b ]:

[ Q^T b begin{bmatrix} 0.4472 0.8944 0 0 end{bmatrix} begin{bmatrix} 1 2 end{bmatrix} begin{bmatrix} 2.2361 0 end{bmatrix} ]

Solve [ R y Q^T b ]:

[ begin{bmatrix} 4.4721 2.2361 0 0 end{bmatrix} begin{bmatrix} y_1 y_2 end{bmatrix} begin{bmatrix} 2.2361 0 end{bmatrix} ]

From this, we can solve for [ y_1 ] and [ y_2 ].

Find [ x ]:

Solve for [ x ] using [ R^{-1} y ].

Conclusion

Using QR factorization as a method for solving systems of linear equations offers several advantages, especially when dealing with least squares problems or when the matrix [ A ] is ill-conditioned. This robust technique simplifies the solution process and ensures more accurate and reliable results.