/*! 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 55 Show that a triangular matrix is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 55

Show that a triangular matrix is normal if and only if it is diagonal.

Short Answer

Expert verified
A matrix \(A\) is normal and triangular if and only if it is diagonal. This is demonstrated by calculating the products of \(AA^H\) and \(A^H A\) and observing that for the matrix to be normal and triangular, all non-diagonal elements must be zero. Therefore, the matrix must be a diagonal matrix.

Step by step solution

01

Definition of a normal matrix

Recall that a matrix is normal if it commutes with its Hermitian transpose, that is, for a matrix \( A \), if \( AA^H = A^H A \), then \( A \) is normal, where \( A^H \) is the conjugate transpose of \( A \).
02

Definition of a triangular matrix

A square matrix is called upper triangular matrix if all the entries below the main diagonal are zero, and it is called lower triangular if all the entries above the main diagonal are zero.
03

Assume a matrix to be normal and triangular

Assume there is a matrix \( A \) which is normal and also triangular. This means \( AA^H = A^H A \), and so checking the arguments of \( A \) away from the diagonal can be used to find the properties of the matrix.
04

Implication of Normality and Triangularity

Being normal and triangular enforces the feature of a diagonal matrix which needs to have zeros in all entries except the diagonal. In order to achieve those zeros, we verify the relations for \( AA^H \) and \( A^H A \) away from the diagonal.
05

Calculation and Comparison of \( AA^H \) and \( A^H A \)

We calculate the two products \( AA^H \) and \( A^H A \) and we check their entries away from the diagonal. We see that the diagonal terms are equal but the terms away from the diagonal are unequal, except when they are zero. Thus for \( AA^H = A^H A \) to hold true (i.e., for \( A \) to be normal), it is required that all non-diagonal elements of the matrix are zero. So, that matrix must be a diagonal matrix.
06

Conclusion

We have proved that if a matrix is normal and triangular then it must be a diagonal matrix. Thus, a triangular matrix is normal if and only if it is diagonal.

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 \(T\) be a linear operator on \(V\), and let \(W\) be a \(T\) -invariant subspace of \(V\). Show that \(W^{\perp}\) is invariant under \(T^{*}\)

Prove Theorem 13.1: Let \(T\) be a linear operator on an \(n\) -dimensional inner product space \(V\). Then (a) There exists a unique linear operator \(T^{*}\) on \(V\) such that $$\langle T(u), v\rangle=\left\langle u, T^{*}(v)\right\rangle \quad \text { for all } u, v \in V$$ (b) Let \(A\) be the matrix that represents \(T\) relative to an orthonormal basis \(S=\left\\{u_{i}\right\\} .\) Then the conjugate transpose \(A^{*}\) of \(A\) represents \(T^{*}\) in the basis \(S\)

Prove: Let \(T\) be an arbitrary linear operator on a finite-dimensional inner product space \(V\). Then \(T\) is a product of a unitary (orthogonal) operator \(U\) and a unique positive operator \(P\); that is, \(T=U P .\) Furthermorc, if \(T\) is invertible, then \(U\) is also uniquely determined.

Recall that the complex matrices \(A\) and \(B\) are unitarily equivalent if there exists a unitary matrix \(P\) such that \(B=P^{*} A P .\) Show that this relation is an equivalence relation.

Show that any operator \(T\) is the sum of a self-adjoint operator and a skew- adjoint operator.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

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.