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Chapter 2: Problem 15

Find the \(\mathrm{QR}\) factorisation of the matrix $$ A=\left(\begin{array}{rr} 9 & -6 \\ 12 & -8 \\ 0 & 20 \end{array}\right) $$ and hence find the least squares solution of the system of linear equations $$ \begin{aligned} 9 x-6 y &=300 \\ 12 x-8 y &=600 \\ 20 y &=900 \end{aligned} $$

Short Answer

Expert verified
QR factorization found with orthonormal columns. Solve \\( R \mathbf{x} = Q^T \mathbf{b} \\) for least squares solution.

Step by step solution

01

Finding Column Vectors

Let's denote the columns of matrix \( A \) as \( \mathbf{a_1} \) and \( \mathbf{a_2} \). This gives us \( \mathbf{a_1} = \begin{pmatrix} 9 \ 12 \ 0 \end{pmatrix} \) and \( \mathbf{a_2} = \begin{pmatrix} -6 \ -8 \ 20 \end{pmatrix} \).
02

Compute Orthogonal Vector \( \mathbf{q_1} \)

To find \( \mathbf{q_1} \), we first calculate the norm of \( \mathbf{a_1} \): \[ \| \mathbf{a_1} \| = \sqrt{9^2 + 12^2 + 0^2} = 15 \]Then, \( \mathbf{q_1} \) is:\[ \mathbf{q_1} = \frac{1}{15} \begin{pmatrix} 9 \ 12 \ 0 \end{pmatrix} = \begin{pmatrix} \frac{3}{5} \ \frac{4}{5} \ 0 \end{pmatrix} \]
03

Compute Orthogonal Vector \( \mathbf{q_2} \)

Project \( \mathbf{a_2} \) onto \( \mathbf{q_1} \) to find the part along \( \mathbf{q_1} \) and then subtract to get the orthogonal component:\[ \mathbf{a_2} = \begin{pmatrix} -6 \ -8 \ 20 \end{pmatrix} \]\[ \text{proj}_{\mathbf{q_1}} \mathbf{a_2} = \left( \mathbf{a_2} \cdot \mathbf{q_1} \right) \mathbf{q_1} \]\( \mathbf{a_2} \cdot \mathbf{q_1} = \frac{-6 \cdot 3}{5} + \frac{-8 \cdot 4}{5} = -\frac{90}{25} = -14.4 \)Then, calculate \( \mathbf{b} = \mathbf{a_2} - \text{proj}_{\mathbf{q_1}} \mathbf{a_2} \):\[ \text{proj}_{\mathbf{q_1}} \mathbf{a_2} = -14.4 \begin{pmatrix} \frac{3}{5} \ \frac{4}{5} \ 0 \end{pmatrix} = \begin{pmatrix} -8.64 \ -11.52 \ 0 \end{pmatrix} \]Thus,\[ \mathbf{b} = \begin{pmatrix} -6 \ -8 \ 20 \end{pmatrix} - \begin{pmatrix} -8.64 \ -11.52 \ 0 \end{pmatrix} = \begin{pmatrix} 2.64 \ 3.52 \ 20 \end{pmatrix} \]Now, normalize \( \mathbf{b} \) to find \( \mathbf{q_2} \):\[ \mathbf{q_2} = \frac{1}{\sqrt{2.64^2 + 3.52^2 + 20^2}} \begin{pmatrix} 2.64 \ 3.52 \ 20 \end{pmatrix} \approx \begin{pmatrix} 0.130 \ 0.173 \ 0.985 \end{pmatrix} \]
04

Construct Matrix \( Q \)

The orthogonal matrix \( Q \) can now be formed using the vectors \( \mathbf{q_1} \) and \( \mathbf{q_2} \):\[ Q = \begin{pmatrix} \frac{3}{5} & 0.130 \ \frac{4}{5} & 0.173 \ 0 & 0.985 \end{pmatrix} \]
05

Construct Matrix \( R \)

Matrix \( R \) is an upper triangular matrix. Compute \( R \) using the relations \( R = Q^T A \):\[ R = \begin{pmatrix} 15 & -14.4 \ 0 & 20.646 \end{pmatrix} \]
06

Solving the Least Squares Problem

Set up the least squares problem: \( A \mathbf{x} \approx \mathbf{b} \). Use factorization \( QR \) to solve for \( \mathbf{x} \) using: \( R \mathbf{x} = Q^T \mathbf{b} \).Convert the system:\[ Q^T \mathbf{b} = \begin{pmatrix} 15 & -14.4 \ 0 & 20.646 \end{pmatrix} \mathbf{x} \]Solve the equations.

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Key Concepts

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

Least Squares Solution
When working with systems of linear equations, one of the key goals is to find solutions that best fit the given data, especially when the system does not have an exact solution. This is where the concept of the least squares solution comes into play. The least squares approach is used to minimize the sum of the squares of the residuals, the differences between the actual and estimated values.In simpler terms, rather than trying to find a perfect solution, the least squares solution seeks to find the values that make the equations most closely aligned with the data provided. These solutions are particularly important in fields like statistics and data analysis.To find the least squares solution using QR factorization:
  • Factorize the matrix as \( A = QR \), where \( Q \) is an orthogonal matrix and \( R \) is an upper triangular matrix.
  • Instead of solving \( A\mathbf{x} = \mathbf{b} \), which is difficult when \( A \) is not full rank or over-determined, solve \( R\mathbf{x} = Q^T\mathbf{b} \).
  • This transformation makes the problem easier, as \( R \) is upper triangular, simplifying the computation.
Remember, the least squares solution is the closest approximation when an exact match isn't possible.
Orthogonal Matrix
Orthogonal matrices are a fascinating and extremely useful concept in linear algebra. An orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors, meaning each column is orthogonal to the others and each has a magnitude of one.One of the most important properties of an orthogonal matrix \( Q \) is that its transpose equals its inverse: \( Q^T = Q^{-1} \). This property greatly simplifies calculations, especially when performing operations like matrix multiplication.Why are orthogonal matrices used in QR factorization?
  • They maintain vector norms, making computations stable and reducing errors in numerical calculations.
  • They simplify the process of finding least squares solutions by decoupling complex relationships.
  • During QR factorization, an orthogonal matrix \( Q \) is paired with an upper triangular matrix \( R \) to form an easily solvable system.
By utilizing orthogonal matrices, solving linear equations becomes more efficient and reliable, significantly benefiting both theoretical and practical applications.
Upper Triangular Matrix
An upper triangular matrix is a type of square matrix where all the elements below the main diagonal are zero. In other words, all the non-zero elements are located on or above the diagonal.In linear algebra, upper triangular matrices are significant because they simplify many matrix calculations. When you have an upper triangular matrix \( R \), working with it to solve linear equations is straightforward because:
  • Solving systems of equations translates into back substitution, a direct method for finding solutions quickly and efficiently.
  • They arise naturally in reduction steps, such as the Gaussian elimination process.
  • Upper triangular matrices maintain interpretation in terms of linear transformations, essential for calculus and other higher-dimensional calculations.
In QR factorization, an upper triangular matrix \( R \) pairs with an orthogonal matrix \( Q \) to form the product \( A = QR \). This setup makes it easy to solve least squares problems as the structure of \( R \) eliminates many possible computational issues and highlights key features of the solution.

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Most popular questions from this chapter

Let \(n \geq 2 .\) Consider a matrix \(A \in \mathbb{R}^{n \times n}\) whose every leading principal submatrix of order less than \(n\) is nonsingular. Show that \(A\) can be factored in the form \(A=L D U\), where \(L \in \mathbb{R}^{n \times n}\) is unit lower triangular, \(D \in \mathbb{R}^{n \times n}\) is diagonal and \(U \in \mathbb{R}^{n \times n}\) is unit upper triangular. If the factorisation \(A=L U\) is known, where \(L\) is unit lowertriangular and \(U\) is upper triangular, show how to find the factors of the transpose \(A^{\mathrm{T}}\).

Let \(n \geq 2\). Given the matrix \(A=\left(a_{i j}\right) \in \mathbb{R}^{n \times n}\), the permutation \(\operatorname{matrix} Q \in \mathbb{R}^{n \times n}\) reverses the order of the rows of \(A\), so that \((Q A)_{i, j}=a_{n+1-i, j}\). If \(L \in \mathbb{R}^{n \times n}\) is a lower triangular matrix, what is the structure of the matrix \(Q L Q\) ? Show how to factorise \(A \in \mathbb{R}^{n \times n}\) in the form \(A=U L\), where \(U \in \mathbb{R}^{n \times n}\) is unit upper triangular and \(L \in \mathbb{R}^{n \times n}\) is lower triangular. What conditions on \(A\) will ensure that the factorisation exists? Give an example of a square matrix \(A\) which cannot be factorised in this way.

Given a matrix \(A \in \mathbb{R}^{n \times n}\), define the matrix \(B \in \mathbb{R}^{n \times 2 n}\) in which the first \(n\) columns are the columns of \(A\), and the last \(n\) columns are the columns of the identity matrix \(I_{n}\). Consider the following computational scheme. Treat the rows of the matrix \(B\) in order, so that \(j=1,2, \ldots, n\). Multiply every element in row \(j\) by the reciprocal of the diagonal element, \(1 / b_{j j}\); then, replace every element \(b_{i k}\) which is not in row \(j\), so that \(i \neq j\), by \(b_{i k}-b_{i j} b_{j k}\). Show that the result is equivalent to multiplying \(B\) on the left by a sequence of matrices. Explain why, at the end of the computation, the first \(n\) columns of \(B\) are the columns of the identity matrix \(I_{n}\), and the last \(n\) columns are the columns of the inverse matrix \(A^{-1}\). Give a condition on the matrix \(A\) which will ensure that the computation does not break down. Show that the process as described requires approximately \(2 n^{3}\) multiplications, but that, if the multiplications in which one of the factors is zero are not counted, the total is approximately \(n^{3}\).

Suppose that for a matrix \(A \in \mathbb{R}^{n \times n}\), $$ \sum_{i=1}^{n}\left|a_{i j}\right| \leq C, \quad j=1,2, \ldots, n . $$ Show that, for any vector \(x \in \mathbb{R}^{n}\), $$ \sum_{i=1}^{n}\left|(A x)_{i}\right| \leq C\|x\|_{1} $$ Find a nonzero vector \(\boldsymbol{x}\) for which equality can be achieved, and deduce that $$ \|A\|_{1}=\max _{j=1}^{n} \sum_{i=1}^{n}\left|a_{i j}\right| $$
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