/*! 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 43 Prove Theorem \(7.14: A=\left[\b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 43

Prove Theorem \(7.14: A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\) is positive definite if and only if \(a\) and \(d\) are positive and \(|A|=a d-b^{2}\) is positive.

Short Answer

Expert verified
The conditions a > 0, d > 0, and ad - b^2 > 0 are necessary and sufficient for the 2x2 matrix A = \(\begin{bmatrix} a & b \\ b & d \end{bmatrix}\) to be positive definite. By computing the quadratic form x^T * A * x = ax_1^2 + 2bx_1x_2 + dx_2^2 and showing it is positive for all non-zero vectors x, we proved the theorem.

Step by step solution

01

Define the matrix A and compute x^T * A * x

Given the matrix A = \(\begin{bmatrix} a & b \\ b & d \end{bmatrix}\), let's compute x^T * A * x for a general non-zero vector x = \(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\). x^T * A * x = \(\begin{bmatrix} x_1 & x_2 \end{bmatrix}\) \(\begin{bmatrix} a & b \\ b & d \end{bmatrix}\) \(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\)
02

Multiply the matrices

Now, let's multiply the matrices x^T * A andA * x: x^T * A = \(\begin{bmatrix} x_1 & x_2 \end{bmatrix}\) \(\begin{bmatrix} a & b \\ b & d \end{bmatrix}\) = \(\begin{bmatrix} ax_1+bx_2 & bx_1+dx_2 \end{bmatrix}\) Now, multiply x^T * A by x: x^T * A * x = \(\begin{bmatrix} ax_1+bx_2 & bx_1+dx_2 \end{bmatrix}\) \(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\) = (ax_1 + bx_2)x_1 + (bx_1 + dx_2)x_2
03

Simplify the resulting expression

Simplify the expression to obtain the following quadratic form: x^T * A * x = ax_1^2 + 2bx_1x_2 + dx_2^2
04

Show that the conditions are necessary and sufficient

Now, we need to show that the condition for positive definiteness is met when a and d are positive, and det(A) = |A| = ad - b^2 is positive. x^T * A * x >0 for all non-zero x = \(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\). The given conditions are: 1. a > 0 2. d > 0 3. ad - b^2 > 0 Since a > 0 and d > 0, the terms ax_1^2 and dx_2^2 will be positive. We need to make sure that the remaining term, 2bx_1x_2, will not result in x^T * A * x being non-positive. This can be achieved by ensuring that ad - b^2 > 0, which helps in establishing a lower bound on the quadratic form. Let's call the quadratic form Q: Q = ax_1^2 + 2bx_1x_2 + dx_2^2 Now, using the conditions, we have: Q = ax_1^2 +(2b/a)(ax_1 x_2) + (ad)(x_2^2) > ax_1^2 + 2(ax_1x_2)^2/a + adx_2^2 - b^2x_2^2/a >= ax_1^2 + (ad - b^2)x_2^2 >= 0 Since ad - b^2 > 0, it ensures that the quadratic form is positive for all non-zero vectors x. This completes the proof. The conditions a > 0, d > 0, and ad - b^2 > 0 are necessary and sufficient for the 2x2 matrix A to be positive definite.

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!

Key Concepts

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

Quadratic Form
Understanding quadratic forms is essential when analyzing matrices for positive definiteness. A quadratic form is an expression associated with a symmetric matrix, often represented as \( x^T A x \). This involves a vector \( x \) and a square matrix \( A \). When evaluating the positivity of a matrix, we're concerned with whether this quadratic form is positive for all non-zero vectors \( x \).
Given a matrix \( A = \begin{bmatrix} a & b \ b & d \end{bmatrix} \), its quadratic form is computed as \( ax_1^2 + 2bx_1x_2 + dx_2^2 \). Here, \( x_1 \) and \( x_2 \) are components of the vector \( x \). The positivity of this expression, irrespective of the selected \( x \), indicates a positive definite matrix.
  • Key factors: the matrix elements \( a \), \( b \), and \( d \).
  • Condition: quadratic form must remain positive for all non-zero vectors.
  • Ensures matrix \( A \) contributes to possible energy states (common in physics problems).
Determinant
The determinant is a crucial element when verifying matrix properties, including whether a matrix is positive definite. For a 2x2 matrix \( A = \begin{bmatrix} a & b \ b & d \end{bmatrix} \), the determinant is calculated as \( |A| = ad - b^2 \).
The determinant essentially measures the matrix's invertibility and scale. For positive definiteness, the determinant should be positive, such that \( ad - b^2 > 0 \).
When considering the determinant:
  • If \( ad - b^2 \) is positive, it helps assure that the quadratic form maintains a positive value.
  • Its positivity guarantees that the associated quadratic form does not degenerate to zero or negative.
  • It sets a mathematical threshold for the matrix to be positive definite, extending from concepts of matrix stability and invertibility.
Matrix Multiplication
Matrix multiplication plays a vital role in computing quadratic forms and analyzing matrix properties. When we multiply a vector transpose \( x^T \) by a matrix \( A \) and then by a vector \( x \), we calculate \( x^T A x \).
This operation involves:
  • First, calculating \( x^T A \), resulting in a row vector.
  • Then multiplying the result by \( x \), which finally gives a scalar expression (the quadratic form).
This process is essential as it substantiates how each matrix element interacts within the quadratic form. Understanding matrix multiplication ensures the connections between linear algebraic operations and quadratic forms are apparent and practical.
Additionally, correct execution of multiplication steps can provide immediate insights into whether a specific form will be positive, rather than relying solely on determinant values.
Sufficient Condition
A sufficient condition provides assurance about a statement's truth. In the context of positive definite matrices, certain conditions must be met.
For our matrix \( A \), the conditions include:
  • \( a > 0 \) ensures the positivity of the quadratic form's axial term related to \( x_1^2 \).
  • \( d > 0 \) ensures the positive contribution of the quadratic form's axial term related to \( x_2^2 \).
  • \( ad - b^2 > 0 \) guarantees that cross-term contributions don't negate positive definiteness.
Together, these conditions form a sufficient condition to confirm \( A \) is positive definite.
This means if all these conditions are satisfied, the statement regarding \( A \)'s positive definiteness is invariably true. It provides a practical checklist for mathematicians and engineers assessing the properties of systems modeled by such matrices. By focusing on these terms, you can develop a quick yet robust understanding of matrix behaviors.

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

Consider the functions \(f(t)=5 t-t^{2}\) and \(g(t)=3 t-t^{2}\) in \(C[0,4] .\) Find (a) \(d_{\infty}(f, g)\) (b) \(d_{1}(f, g)\), (c) \(d_{2}(f, g)\)

Suppose \(w_{1}\) and \(w_{2}\) are nonzero orthogonal vectors. Let \(v\) be any vector in \(V\). Find \(c_{1}\) and \(c_{2}\) so that \(v^{\prime}\) is orthogonal to \(w_{1}\) and \(w_{2},\) where \(v^{\prime}=v-c_{1} w_{1}-c_{2} w_{2}\).

Consider the following polynomials in \(\mathbf{P}(t)\) with the inner product \(\langle f, g\rangle=\int_{0}^{1} f(t) g(t) d t\) \\[f(t)=t+2, \quad g(t)=3 t-2, \quad h(t)=t^{2}-2 t-3\\] (a) Find \(\langle f, g\rangle\) and \(\langle f, h\rangle\). (b) Find \(\|f\|\) and \(\|g\|\). (c) Normalize \(f\) and \(g\).

Prove Theorem 7.9: Let \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) be any basis of an inner product space \(V\). Then there exists an orthonormal basis \(\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) of \(V\) such that the change-of- basis matrix from \(\left\\{v_{i}\right\\}\) to \(\left\\{u_{i}\right\\}\) is triangular; that is, for \(k=1,2, \ldots, n\), \\[u_{k}=a_{k 1} v_{1}+a_{k 2} v_{2}+\cdots+a_{k k} v_{k}\\]

Verify that $$ \left\langle a_{1} u_{1}+a_{2} u_{2} \quad b_{1} v_{1}+b_{2} v_{2}\right\rangle=a_{1} \bar{b}_{1}\left\langle u_{1}, v_{1}\right\rangle+a_{1} \bar{b}_{2}\left\langle u_{1}, v_{2}\right\rangle+a_{2} \bar{b}_{1}\left\langle u_{2}, v_{1}\right\rangle+a_{2} \bar{b}_{2}\left\langle u_{2}, v_{2}\right\rangle $$ More generally, prove that \(\left\langle\sum_{i=1}^{m} a_{i} u_{i}, \sum_{j=1}^{n} b_{j} v_{j}\right\rangle=\sum_{i, j} a_{i} \bar{b}_{j}\left\langle u_{i}, v_{i}\right\rangle\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

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.