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Chapter 5: Problem 18

The columns of \(Q\) were obtained by applying the Gram-Schmidt Process to the columns of \(A\). Find the upper triangular matrix \(R\) such that \(A=Q R\). $$A=\left[\begin{array}{rr} 1 & 3 \\ 2 & 4 \\ -1 & -1 \\ 0 & 1 \end{array}\right], Q=\left[\begin{array}{rc} 1 / \sqrt{6} & 1 / \sqrt{3} \\ 2 / \sqrt{6} & 0 \\ -1 / \sqrt{6} & 1 / \sqrt{3} \\ 0 & 1 / \sqrt{3} \end{array}\right]$$

Short Answer

Expert verified
The upper triangular matrix \(R\) is \[R = \begin{bmatrix} \sqrt{6} & \sqrt{6} \\ 0 & \sqrt{3} \end{bmatrix}.\]

Step by step solution

01

Understand the Given Matrices

You are given two matrices, \[ A = \begin{bmatrix} 1 & 3 \ 2 & 4 \ -1 & -1 \ 0 & 1 \end{bmatrix} \] and \[ Q = \begin{bmatrix} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \ \frac{2}{\sqrt{6}} & 0 \ -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \ 0 & \frac{1}{\sqrt{3}} \end{bmatrix}. \] The matrix \(A\) is derived using the Gram-Schmidt process, resulting in an orthogonal matrix \(Q\) and an upper triangular matrix \(R\), such that \(A = QR\). Our goal is to find \(R\).
02

Express A in Terms of Q and R

Since \(A = QR\), if \(Q\) is given, we can express \(R\) as \(R = Q^T A\). Therefore, to find \(R\), you need to calculate the transpose of \(Q\) and then multiply it by \(A\).
03

Calculate the Transpose of Q

The transpose of matrix \(Q\) is obtained by interchanging its rows and columns:\[ Q^T = \begin{bmatrix} \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & 0 \ \frac{1}{\sqrt{3}} & 0 & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{bmatrix}. \]
04

Multiply Q^T by A

Perform the multiplication \(Q^T A\). For \(R\), calculate:For the first column of \(R\):\[ R_{11} = \frac{1}{\sqrt{6}} \times 1 + \frac{2}{\sqrt{6}} \times 2 + -\frac{1}{\sqrt{6}} \times (-1) + 0 \times 0 = \frac{1}{\sqrt{6}}(1 + 4 + 1) = \sqrt{6}. \]For the second column of \(R\):\[ R_{12} = \frac{1}{\sqrt{6}} \times 3 + \frac{2}{\sqrt{6}} \times 4 + -\frac{1}{\sqrt{6}} \times (-1) + 0 \times 1 = \frac{1}{\sqrt{6}}(3 + 8 + 1) = \sqrt{6}. \]For \(R_{22}\):\[ R_{22} = \frac{1}{\sqrt{3}} \times 3 + 0 \times 4 + \frac{1}{\sqrt{3}} \times (-1) + \frac{1}{\sqrt{3}} \times 1 = \frac{1}{\sqrt{3}} \times 3 = \sqrt{3}. \]
05

Form the Upper Triangular Matrix R

Combine the calculated elements into the matrix \(R\):\[ R = \begin{bmatrix} \sqrt{6} & \sqrt{6} \ 0 & \sqrt{3} \end{bmatrix}. \]\(R\) is an upper triangular matrix derived from our calculations, confirming that \(A = QR\).

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Key Concepts

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

Orthogonal Matrix
An orthogonal matrix is a special kind of square matrix whose columns and rows are orthogonal unit vectors. In simpler terms, this means that the matrix's columns and rows are perpendicular to each other and have a length of one. This property makes them particularly useful in various mathematical computations and transformations. Orthogonal matrices have some interesting attributes:
  • The inverse of an orthogonal matrix is equal to its transpose. This means if you need the inverse, you just flip the matrix over its diagonal.
  • Orthogonal matrices preserve the length of vectors. When you multiply a vector by an orthogonal matrix, the vector's length remains unchanged.
  • The determinant of an orthogonal matrix is always \(\pm 1\).
In our given problem, the matrix \(Q\) represents an orthogonal matrix derived using the Gram-Schmidt process on matrix \(A\). This is crucial because it allows us to decompose \(A\) into a product of \(Q\) and an upper triangular matrix \(R\).
Upper Triangular Matrix
An upper triangular matrix is a type of matrix where all the entries below the main diagonal are zero. In mathematical terms, for an upper triangular matrix \(R\), the elements \(R_{ij}\) are zero for all \(i > j\). These matrices are significant in linear algebra for a few reasons:
  • It simplifies solving systems of linear equations, especially during back substitution.
  • They are easier to compute the determinant and inverse for, compared to other matrix forms.
  • Any square matrix can be decomposed into a product that includes an upper triangular matrix through processes like LU decomposition or QR factorization.
In the context of the original problem, the calculation of the matrix \(R\) as an upper triangular form confirmed that \(A = QR\). This highlights the relationships and transformations matrices can undergo in mathematical processes.
Matrix Multiplication
Matrix multiplication is the process of multiplying two matrices to produce a new matrix. Unlike regular multiplication, it's important to note that matrix multiplication is not commutative. This means that \(AB\) does not necessarily equal \(BA\).Here鈥檚 a simple guide to multiplying matrices:
  • Ensure the number of columns in the first matrix is equal to the number of rows in the second matrix.
  • The resulting matrix will have dimensions equal to the number of rows of the first matrix and the number of columns of the second matrix.
  • To find each element in the resulting matrix, multiply the corresponding entries and then sum them up.
In our solution, the multiplication of \(Q^T\) and \(A\) was necessary to find the matrix \(R\). This illustrates how understanding matrix multiplication, with its particular rules and conditions, enables us to solve problems like determining \(R\) in the equation \(A=QR\).
Transpose of a Matrix
The transpose of a matrix is a new matrix obtained by switching the rows and columns of the original matrix. If you have a matrix \(M\), its transpose, denoted as \(M^T\), is formed by turning the first row of \(M\) into the first column of \(M^T\), the second row of \(M\) into the second column, and so on.Transposing matrices has useful properties:
  • The transpose of a transpose gives you back the original matrix, i.e., \((M^T)^T = M\).
  • The transpose of a product of matrices is the product of their transposes in reverse order, i.e., \((AB)^T = B^T A^T\).
  • Linearity: Transpose maintains addition and scalar multiplication, so \((aM+bN)^T = aM^T+bN^T\).
In the original exercise, the transpose of the orthogonal matrix \(Q\) was needed to compute \(R\). This step demonstrates the importance of being able to manipulate matrices through simple operations like transposition in order to achieve the desired results in linear algebra.

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