/*! 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 10 Let \(A\) be a singular \(n \tim... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 10

Let \(A\) be a singular \(n \times n\) matrix. Show that \(A^{T} A\) is positive semidefinite, but not positive definite.

Short Answer

Expert verified
In conclusion, given a singular \(n \times n\) matrix \(A\), \(A^{T} A\) is positive semidefinite because the quadratic form \(x^{T}A^{T}Ax = y^T y \geq 0\). However, \(A^{T} A\) is not positive definite because there exists a non-zero vector \(x\) with \(Ax = 0\) such that \(x^{T}A^{T}Ax = 0\).

Step by step solution

01

Show that \(A^{T} A\) is positive semidefinite

Given a singular \(n \times n\) matrix \(A\), we need to show that \(A^{T} A\) is positive semidefinite. To do this, we will examine the quadratic form \(x^{T}A^{T}Ax\) for any non-zero vector \(x\): \[x^{T}A^{T}Ax\] Notice that if we define a new vector \(y = Ax\), then we can rewrite the quadratic form as: \[x^{T}A^{T}Ax = x^{T}y^Ty = y^{T}y\] Now we know that \(y^Ty\) is the sum of squares of the elements of \(y\): \[ y^T y = y_1^2 + y_2^2 + \dots + y_n^2 \] Since the sum of squares is always non-negative, it follows that: \[ x^{T}A^{T}Ax = y^T y \geq 0\] Thus, \(A^{T} A\) is positive semidefinite.
02

Show that \(A^{T} A\) is not positive definite

As we know that \(A\) is a singular matrix, this implies that its determinant is 0: \[ \det(A) = 0 \] This means that \(A\) has at least one non-trivial (non-zero) solution in its null space. So, there exists a non-zero vector \(x\) such that: \[ Ax = 0 \] Now, let's examine the quadratic form for this specific vector \(x\): \[ x^{T}A^{T}Ax \] As \(Ax = 0\), we can replace \(Ax\) by \(0\): \[ x^{T}A^{T}Ax = x^{T}0 = 0 \] So, we have found a non-zero vector \(x\) for which the quadratic form \(x^{T}A^{T}Ax = 0\). Hence, \(A^{T} A\) is not positive definite. In conclusion, \(A^{T} A\) is positive semidefinite, but not positive definite for a singular \(n \times n\) matrix \(A\).

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

Find the matrix associated with each of the following quadratic forms: (a) \(3 x^{2}-5 x y+y^{2}\) (b) \(2 x^{2}+3 y^{2}+z^{2}+x y-2 x z+3 y z\) (c) \(x^{2}+2 y^{2}+z^{2}+x y-2 x z+3 y z\)

Show that the diagonal entries of a Hermitian matrix must be real.

Let \(\mathbf{x}, \mathbf{y}\) be nonzero vectors in \(\mathbb{R}^{n}, n \geq 2,\) and let \(A=\mathbf{x y}^{T} .\) Show that (a) \(\lambda=0\) is an eigenvalue of \(A\) with \(n-1\) linearly independent eigenvectors and consequently has multiplicity at least \(n-1\) (see Exercise 16 (b) the remaining eigenvalue of \(A\) is \\[ \lambda_{n}=\operatorname{tr} A=\mathbf{x}^{T} \mathbf{y} \\] and \(\mathbf{x}\) is an eigenvector belonging to \(\lambda_{n}\) (c) if \(\lambda_{n}=\mathbf{x}^{T} \mathbf{y} \neq 0,\) then \(A\) is diagonalizable.

Show that \(e^{A}\) is nonsingular for any diagonalizable matrix \(A\)

Show that if \(A\) is a normal matrix, then each of the following matrices must also be normal: (a) \(A^{H}\) (b) \(I+A\) (c) \(A^{2}\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics 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.