Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let \(P\) be an orthogonal matrix. a. Show that \(\operatorname{det} P=1\) or \(\operatorname{det} P=-1\). b. Give \(2 \times 2\) examples of \(P\) such that \(\operatorname{det} P=1\) and \(\operatorname{det} P=-1\) c. If \(\operatorname{det} P=-1\), show that \(I+P\) has no inverse. \(\left[\right.\) Hint \(\left.: P^{T}(I+P)=(I+P)^{T} .\right]\) d. If \(P\) is \(n \times n\) and det \(P \neq(-1)^{n}\), show that \(I-P\) has no inverse. \(\left[\right.\) Hint: \(\left.P^{T}(I-P)=-(I-P)^{T} .\right]\)

Short Answer

Expert verified
Orthogonal matrices have \(\operatorname{det} P = \pm 1\). Examples: \(\operatorname{det} P = 1\) for identity, \(-1\) for permutation matrix. \(I+P\) is non-invertible for \(\operatorname{det} P = -1\), \(I-P\) is non-invertible if \(\operatorname{det} P \neq (-1)^n\).

Step by step solution

01

Definition of an Orthogonal Matrix

An orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors. This means that for a matrix \( P \), it satisfies \( P^T P = I \) where \( P^T \) is the transpose of \( P \) and \( I \) is the identity matrix.
02

Determinant Property of Orthogonal Matrices

For an orthogonal matrix \( P \), the determinant has the property that \( \det(P^T) = \det(P) \) and \( \det(P^T P) = \det(I) = 1 \). This implies \( \det(P)^2 = 1 \), leading to \( \det(P) = \pm 1 \).
03

Examples of 2x2 Orthogonal Matrices

Consider the matrix \( P = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). It is orthogonal and \( \det(P) = 1 \). Another example, \( P = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), is also orthogonal and \( \det(P) = -1 \).
04

Showing \( I+P \) is Non-Invertible When \( \det(P) = -1 \)

If \( \det(P) = -1 \), consider \( P^T (I+P) = (I+P)^T \). Since \( P^T = P^{-1} \) for orthogonal matrices, this implies \( P^{-1}(I+P) = (I+P)^T \). Taking the determinant, \( \det(P^{-1})\det(I+P) = \det((I+P)^T) \). But if \( \det(I+P) eq 0 \), it should lead to an inconsistency with \( \det(P) = -1 \), thus \( I+P \) is singular.
05

Condition for \( I-P \) Non-Invertibility

Given \( P^T (I-P) = -(I-P)^T \), and det \( P eq (-1)^n \), consider \( (I-P)^T = -(I-P) \). If \( I-P \) were invertible, we can multiply both sides by this inverse leading to a contradiction unless \( det(I-P) = 0 \), which signifies that \( I-P \) is non-invertible.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Determinant
The determinant is a special number that can be calculated from a square matrix. It helps us understand many properties of the matrix, such as whether it is invertible. For an orthogonal matrix \( P \), an interesting property holds: the determinant can only be \( 1 \) or \( -1 \).

Why is this the case? Since \( P \) is orthogonal, it satisfies \( P^T P = I \), where \( P^T \) is the transpose of \( P \) and \( I \) is the identity matrix. When we take the determinant of both sides, we get \( \det(P^T P) = \det(I) = 1 \). This results in \( \det(P)^2 = 1 \), leading to \( \det(P) = \pm 1 \).

Determinants are crucial for understanding matrices because they provide insight into the matrix's characteristics, such as rotation and reflection in geometry.
Matrix Invertibility
Matrix invertibility refers to the ability to find a matrix's inverse. A matrix is invertible if there is another matrix that, when multiplied with it, yields the identity matrix. However, not all matrices have an inverse.

For example, if an orthogonal matrix \( P \) has \( \det(P) = 1 \), it is usually invertible. However, if \( \det(P) = -1 \), things change. In the case of \( I+P \), where \( \det(P) = -1 \), the matrix is non-invertible because the resulting determinant calculation leads to an inconsistency, showing \( \det(I+P) \) equals zero. Similarly, if \( \det(P) eq (-1)^n \) for an \( n \times n \) matrix, \( I-P \) also becomes non-invertible.

This concept helps in determining when a matrix can perform operations that rely on its inverse.
2x2 Matrices
2x2 matrices are a simple yet powerful form of matrices that often serve as introductory examples for learning matrix concepts. A 2x2 matrix consists of two rows and two columns, allowing various operations and applications.

For orthogonal 2x2 matrices, we can have:\[ P = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]
This matrix has a determinant of \( 1 \) and represents identity or no change.

Another example is:\[ P = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \]
This matrix swaps the values and has a determinant of \( -1 \).

These examples show how 2x2 matrices can represent basic geometric transformations like identity and reflections.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(U\) is a subspace of \(\mathbb{R}^{n}\), show that \(\operatorname{proj}_{U} \mathbf{x}=\mathbf{x}\) for all \(\mathbf{x}\) in \(U\)

Let \(A\) be positive definite and write \(d_{r}=\operatorname{det}^{(r)} A\) for each \(r=1,2, \ldots, n .\) If \(U\) is the upper triangular matrix obtained in step 1 of the algorithm, show that the diagonal elements \(u_{11}, u_{22}, \ldots, u_{n n}\) of \(U\) are given by \(u_{11}=d_{1}, u_{j j}=d_{j} / d_{j-1}\) if \(j>1\). [Hint: If \(L A=U\) where \(L\) is lower triangular with \(1 \mathrm{~s}\) on the diagonal, use block multiplication to show that \(\operatorname{det}^{(r)} A=\operatorname{det}^{(r)} U\) for each \(\left.r .\right]\)

a. If \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\), show that \(U^{-1} A U\) is not diagonal for any invertible complex matrix \(U\). b. If \(A=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]\), show that \(U^{-1} A U\) is not upper triangular for any real invertible matrix \(U\).

Let \(C\) be a binary linear \(n\) -code (over \(\mathbb{Z}_{2}\) ). Show that either each word in \(C\) has even weight, or half the words in \(C\) have even weight and half have odd weight. [Hint: The dimension theorem.]

In each case find the QR-factorization of \(A\). a. \(A=\left[\begin{array}{rr}1 & -1 \\ -1 & 0\end{array}\right]\) b. \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 1\end{array}\right]\) c. \(A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\) d. \(A=\left[\begin{array}{rrr}1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & -1 & 0\end{array}\right]\)

See all solutions

Recommended explanations on Economics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free