/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Prove that the transpose of an o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 11

Prove that the transpose of an orthogonal matrix is an orthogonal matrix.

Short Answer

Expert verified
We are given that A is an orthogonal matrix, i.e., A * A岬 = I. To prove A岬 is also orthogonal, we first find the transpose of A岬, denoted as (A岬)岬, which is equal to A. We need to show that A岬 * A = I. Since A is orthogonal, taking the transpose of A * A岬 = I, we obtain (A * A岬)岬 = I岬. Using the property (AB)岬 = B岬 * A岬, we rewrite it as (A岬)岬 * A岬 = I. Substituting (A岬)岬 = A, we have A * A岬 = I, proving that A岬 is orthogonal.

Step by step solution

01

Understanding Orthogonal Matrices

: A matrix A is orthogonal if its transpose A岬 is equal to its inverse A鈦宦. Mathematically, this can be written as: A * A岬 = A岬 * A = I where I is the identity matrix.
02

Finding the Transpose of the Transpose Matrix

: We are given that A is an orthogonal matrix, and we are asked to prove that A岬 is also orthogonal. To do this, let's first find the transpose of A岬, which can be denoted as (A岬)岬. Recall that when transposing a matrix, we essentially swap the rows and columns of the matrix. So, when finding the transpose of A岬, we are, in effect, undoing the transposition of A. Therefore, (A岬)岬 = A
03

Verifying the Orthogonality of A岬

: We now need to show that A岬 * A = I. If we can demonstrate that, then we have shown that A岬 is indeed orthogonal. Since A is orthogonal, we have: A * A岬 = I Taking the transpose of both sides of the equation, we obtain: (A * A岬)岬 = I岬 Since the transpose of the identity matrix is the identity matrix, we get: I岬 = I Now, using the property (AB)岬 = B岬 * A岬, we can rewrite the left side of our equation as: (A岬)岬 * A岬 = I But from Step 2, we know that (A岬)岬 = A, so substituting this in our equation, we get: A * A岬 = I Thus, we have proven that A岬 is, indeed, an orthogonal matrix.

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 91影视!

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

Let \(U\) and \(V\) be subspaces of a vector space \(W\) Show that if \(W=U \oplus V,\) then \(U \cap V=\\{0\\}\)

Prove: If \(A\) is an \(m \times n\) matrix and \(\mathbf{x} \in \mathbb{R}^{n},\) then cither \(A \mathbf{x}=0\) or there exists \(\mathbf{y} \in R\left(A^{T}\right)\) such that \(\mathbf{x}^{T} \mathbf{y} \neq 0 .\) Draw a picture similar to Figure 5.2 .2 to illustrate this result geometrically for the case where \(N(A)\) is a two-dimensional subspace of \(\mathbb{R}^{3}\)

Let \(\mathbf{x}\) and \(\mathbf{y}\) be vectors in \(\mathbb{R}^{n}\) and define $$\mathbf{p}=\frac{\mathbf{x}^{T} \mathbf{y}}{\mathbf{y}^{T} \mathbf{y}} \mathbf{y} \quad \text { and } \quad \mathbf{z}=\mathbf{x}-\mathbf{p}$$ (a) Show that \(\mathbf{p} \perp\) z. Thus, \(\mathbf{p}\) is the vector projection of \(\mathbf{x}\) onto \(\mathbf{y} ;\) that is, \(\mathbf{x}=\mathbf{p}+\mathbf{z},\) where \(\mathbf{p}\) and \(\mathbf{z}\) are orthogonal components of \(\mathbf{x},\) and \(\mathbf{p}\) is a scalar multiple of \(\mathbf{y}\) (b) If \(\|\mathbf{p}\|=6\) and \(\|\mathbf{z}\|=8,\) determine the value of \(\|\mathbf{x}\|\)

If \(V\) is an inner product space, show that $$\|\mathbf{v}\|=\sqrt{\langle\mathbf{v}, \mathbf{v}\rangle}$$ satisfies the first two conditions in the definition of a norm.

Show that, for any \(\mathbf{u}\) and \(\mathbf{v}\) in a normed vector space, $$\|\mathbf{u}+\mathbf{v}\| \geq |\|\mathbf{u}\|-\|\mathbf{v}\|$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation
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.