/*! 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 15 Show that the product of two ort... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 15

Show that the product of two orthogonal matrices is also an orthogonal matrix. Is the product of two permutation matrices a permutation matrix? Explain.

Short Answer

Expert verified
The product of two orthogonal matrices is also an orthogonal matrix, as proven by showing both \((AB)^T(AB) = I\) and \(AB(AB)^T = I\) for orthogonal matrices \(A\) and \(B\). Additionally, the product of two permutation matrices is also a permutation matrix, as demonstrated by confirming that the resulting matrix has exactly one '1' in each row and column, and all other elements are '0'.

Step by step solution

01

An orthogonal matrix is a square matrix \(A\) such that its transpose \(A^T\) is equal to its inverse \(A^{-1}\). In other words, \(A^T A = AA^T = I\), where \(I\) denotes the identity matrix. #2: Properties of Orthogonal Matrices and Approach to Prove Their Product is Orthogonal#

In order to prove that the product of two orthogonal matrices is also an orthogonal matrix, we'll take two orthogonal matrices, say \(A\) and \(B\), and prove that the product \(C = AB\) is also orthogonal, i.e., \((AB)^T(AB) = (AB)(AB)^T = I\). #3: Calculation of Transpose of Product Matrix#
02

First, let's find the transpose of the product matrix, i.e., \((AB)^T\). Recall that the transpose of a product of matrices is equal to the product of their transposes in reverse order. So, \((AB)^T = B^T A^T\). #4: Prove Product of Transpose and Product Matrix is Identity Matrix (Left Side)#

Now, let's multiply the transpose \((AB)^T\) with the product matrix \(AB\). \((AB)^T(AB) = (B^T A^T)(AB)\) Using the associative property of matrix multiplication, \((B^T A^T)(AB) = B^T (A^T A) B\) Since \(A\) is orthogonal, \(A^T A = I\). Thus, \(B^T I B = B^T B\) Now, as \(B\) is also orthogonal, \(B^T B = I\). Therefore, we have \((AB)^T(AB) = I\). #5: Prove Product of Product Matrix and Transpose is Identity Matrix (Right side)#
03

Next, let's multiply the product matrix, \(AB\) with its transpose \((AB)^T\). \((AB)(AB)^T = AB(B^T A^T)\) Using the associative property of matrix multiplication, \(AB(B^T A^T) = A(B B^T) A^T\) Since \(B\) is orthogonal, \(B B^T = I\). Thus, \(A I A^T = A A^T\) Now, as \(A\) is also orthogonal, \(A A^T = I\). Therefore, we have \(AB(AB)^T = I\). #6: Conclude Proof for Orthogonal Matrices #

Since we have proven both \((AB)^T(AB) = I\) and \(AB(AB)^T = I\), we can now conclude that the product of two orthogonal matrices is also an orthogonal matrix. #7: Definition of Permutation Matrix #
04

A permutation matrix is a square matrix obtained from the identity matrix by permuting its rows or columns. It has exactly one '1' in each row and each column, and all other elements are '0'. #8: Properties of Permutation Matrices and Approach to Prove Their Product is a Permutation Matrix#

In order to prove that the product of two permutation matrices is a permutation matrix, we'll take two permutation matrices, say \(P\) and \(Q\), and show that their product \(R = PQ\) is also a permutation matrix, i.e., it has exactly one '1' in each row and each column, and all other elements are '0'. #9: Prove Product of Permutation Matrices is a Permutation Matrix#
05

Let's consider the matrices \(P\) and \(Q\) to be \(n \times n\) permutation matrices. Consider the product matrix \(R = PQ\). Let's analyze a general element of \(R\), denoted by \(r_{ij}\). By the definition of matrix multiplication, \(r_{ij}\) is given by the dot product of the \(i\)-th row of \(P\) and the \(j\)-th column of \(Q\). Since each row of \(P\) and each column of \(Q\) has exactly one '1' and the rest are '0's, it is guaranteed that \(r_{ij}\) will be '0' or '1'. Now, let's consider the case when \(r_{ij} = 1\). This means that, for some \(k\) in the range \(1, 2, ..., n\), we have \(p_{ik} = 1\) and \(q_{kj} = 1\). This implies that the \(k\)-th element in the \(i\)-th row of \(P\) and the \(j\)-th column of \(Q\) is '1'. All the other elements in the row and column will be '0'. Therefore, for all rows and columns of \(R\), there is exactly one '1', and the rest are '0's. This confirms that the product of two permutation matrices, \(R = PQ\), is also a permutation matrix. #10: Conclusion#

We have proven that: 1. The product of two orthogonal matrices is also an orthogonal matrix. 2. The product of two permutation matrices is a permutation matrix.

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Most popular questions from this chapter

Consider the vector space \(C[-1,1]\) with inner product defined by $$\langle f, g\rangle=\int_{-1}^{1} f(x) g(x) d x$$ Find an orthonormal basis for the subspace spanned by \(1, x,\) and \(x^{2}\)

Let \(A\) be an \(m \times n\) matrix of rank \(n\) and let \(\mathbf{b} \in \mathbb{R}^{m}\) Show that if \(Q\) and \(R\) are the matrices derived from applying the Gram-Schmidt process to the column vectors of \(A\) and $$\mathbf{p}=c_{1} \mathbf{q}_{1}+c_{2} \mathbf{q}_{2}+\cdots+c_{n} \mathbf{q}_{n}$$ is the projection of \(\mathbf{b}\) onto \(R(A),\) then: (a) \(\mathbf{c}=Q^{T} \mathbf{b}\) (b) \(\mathbf{p}=Q Q^{T} \mathbf{b}\) (c) \(Q Q^{T}=A\left(A^{T} A\right)^{-1} A^{T}\)

Let \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) be an orthonormal basis for an inner product space \(V\) and let $$\mathbf{u}=\mathbf{u}_{1}+2 \mathbf{u}_{2}+2 \mathbf{u}_{3} \quad \text { and } \quad \mathbf{v}=\mathbf{u}_{1}+7 \mathbf{u}_{3}$$ Determine the value of each of the following: (a) \(\langle\mathbf{u}, \mathbf{v}\rangle\) (b) \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\) (c) The angle \(\theta\) between \(\mathbf{u}\) and \(\mathbf{v}\)

Which of the following sets of vectors form an orthonormal basis for \(\mathbb{R}^{2} ?\) (a) \(\left\\{(1,0)^{T},(0,1)^{T}\right\\}\) (b) \(\left\\{\left(\frac{3}{5}, \frac{4}{5}\right)^{T},\left(\frac{5}{13}, \frac{12}{13}\right)^{T}\right\\}\) (c) \(\left\\{(1,-1)^{T},(1,1)^{T}\right\\}\) \((\mathbf{d})\left\\{\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)^{T},\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)^{T}\right\\}\)

For each of the following, use the Gram-Schmidt process to find an orthonormal basis for \(R(A)\) (a) \(A=\left(\begin{array}{rr}-1 & 3 \\ 1 & 5\end{array}\right)\) (b) \(A=\left(\begin{array}{rr}2 & 5 \\ 1 & 10\end{array}\right)\)
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