/*! 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 2 Show that the eigenvalues of a t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 2

Show that the eigenvalues of a triangular matrix are the diagonal elements of the matrix.

Short Answer

Expert verified
In order to show that the eigenvalues of a triangular matrix are the diagonal elements of the matrix, we first calculate the determinant of the matrix (A - 位I) and form the characteristic equation. Since (A - 位I) is a triangular matrix, its determinant is the product of its diagonal elements: \(\det(A - 位I) = (a_{11} - 位)(a_{22} - 位) \cdots (a_{nn} - 位)\). Solving the characteristic equation, the n eigenvalues 位 correspond to the diagonal elements of matrix A: \(位_1 = a_{11}\), \(位_2 = a_{22}\), ... , \(位_n = a_{nn}\). This holds for both upper and lower triangular matrices.

Step by step solution

01

Define a Triangular Matrix

A triangular matrix is a square matrix in which all the entries above (or below) the main diagonal are zero. Here, we can consider an upper triangular matrix as an example, but the same method applies to a lower triangular matrix. Let A be an upper triangular matrix of order n. \[ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{bmatrix} \]
02

Calculate the Characteristic Equation

The eigenvalues of a matrix can be found by solving its characteristic equation. The characteristic equation is given by the determinant of the matrix (A - 位I), where 位 is an eigenvalue and I is the identity matrix of the same order as A. The matrix (A - 位I) is: \[(A - 位I) = \begin{bmatrix} a_{11} - 位 & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} - 位 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} - 位 \end{bmatrix} \] Now, we will calculate the determinant of (A - 位I).
03

Calculate the Determinant

Since the matrix (A - 位I) is a triangular matrix, its determinant is the product of the diagonal elements: \(\det(A - 位I) = (a_{11} - 位)(a_{22} - 位) \cdots (a_{nn} - 位)\)
04

Solve the Characteristic Equation

Now, we need to solve the characteristic equation for 位, which is: \(\det(A - 位I) = (a_{11} - 位)(a_{22} - 位) \cdots (a_{nn} - 位) = 0\) From this equation, we can observe that the n eigenvalues 位 are the diagonal elements of the matrix A, i.e., \(位_1 = a_{11}\), \(位_2 = a_{22}\), ... , \(位_n = a_{nn}\).
05

Conclusion

We have shown that the eigenvalues of a triangular matrix are the diagonal elements of the matrix by solving its characteristic equation and finding its determinant. This is valid for both upper and lower triangular matrices.

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 \(A\) and \(C\) be matrices in \(\mathbb{C}^{m \times n}\) and let \(B \in \mathbb{C}^{n \times r}\) Prove each of the following rules: (a) \(\left(A^{H}\right)^{H}=A\) (b) \((\alpha A+\beta C)^{H}=\bar{\alpha} A^{H}+\bar{\beta} C^{H}\) (c) \(\quad(A B)^{H}=B^{H} A^{H}\)

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) \(A^{2}\)

Show that if \(B\) is a symmetric nonsingular matrix, then \(B^{2}\) is positive definite.

Let \(A\) be an \(n \times n\) matrix with an eigenvalue \(\lambda\) of multiplicity \(n .\) Show that \(A\) is diagonalizable if and only if \(A=\lambda I\)

It follows from Exercise 14 that, for a diagonalizable matrix, the number of nonzero eigenvalues (counted according to multiplicity) equals the rank of the matrix. Give an example of a defective matrix whose rank is not equal to the number of nonzero eigenvalues.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

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.