/*! 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 1 Let \(A\) be an invertible hermi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 1

Let \(A\) be an invertible hermitian matrix. Show that \(A^{-1}\) is hermitian.

Short Answer

Expert verified
To prove that the inverse of a Hermitian matrix is also Hermitian, we start from the inverse property \(AA^{-1} = I\), take the conjugate transpose of both sides, and apply properties of Hermitian matrices and conjugate transposes. We ultimately obtain the equation \((A^{-1})^H A = I\), and by comparing it to the original inverse property equation, we deduce that \((A^{-1})^H = A^{-1}\), which means \(A^{-1}\) is Hermitian.

Step by step solution

01

Write the inverse property

Given that \(A\) is an invertible matrix, we have the property \(AA^{-1} = I\), where \(I\) is the identity matrix.
02

Take the conjugate transpose of both sides

To find the conjugate transpose of the given equation, we can apply the fact that the conjugate transpose of the product of two matrices is equal to the product of the conjugate transposes of the individual matrices in reverse order: \((AB)^H = B^H A^H\). Thus, taking the conjugate transpose of both sides, we get: \((AA^{-1})^H = I^H\) By applying the rule mentioned above, this simplifies to: \((A^{-1})^H A^H = I^H\)
03

Use the Hermitian property of A

Since \(A\) is a Hermitian matrix, we have \(A = A^H\). Also, we know that the conjugate transpose of the identity matrix is equal to itself: \(I^H = I\). Substituting these into the expression obtained in step 2, we get: \((A^{-1})^H A = I\)
04

Compare the two expressions

Compare the expressions from Step 1 and Step 3: \(AA^{-1} = I\) \((A^{-1})^H A = I\) Notice that both equations have the form of matrices multiplying its inverse to form an identity matrix. Since \(A\) and \((A^{-1})^H\) both multiply with \(A^{-1}\) and \(A\), respectively, to obtain the identity matrix, it implies that: \((A^{-1})^H = A^{-1}\)
05

Conclusion

Since we have shown that \((A^{-1})^H = A^{-1}\), we can conclude that the inverse of a Hermitian matrix is also Hermitian.

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.

Inverse of a Matrix
Understanding the inverse of a matrix is fundamental in linear algebra and is particularly relevant when dealing with Hermitian matrices. An inverse of a matrix, denoted as A^{-1}, is a matrix that, when multiplied with the original matrix A, yields the identity matrix I. The equation AA^{-1} = I represents this fundamental relationship.

It's important to note that not all matrices have an inverse; a matrix must be square (same number of rows and columns) and non-singular (having a non-zero determinant) to possess an inverse. When we're given an invertible Hermitian matrix, as in the exercise, we can employ several properties of matrix operations to demonstrate that its inverse will also retain the Hermitian quality, which leads to deeper insights in quantum mechanics, signal processing, and other fields.
Conjugate Transpose
The conjugate transpose of a matrix, often denoted by A^H or A^*, is a critical concept when working with complex matrices such as Hermitian matrices. It is achieved by taking the transpose of the matrix 鈥� swapping rows with columns 鈥� followed by taking the complex conjugate of each element. This operation is essential in the context of Hermitian matrices because for any Hermitian matrix A, it holds that A = A^H.

In the provided exercise, we saw that taking the conjugate transpose of the product of two matrices results in the reverse order of their conjugate transposes, meaning (AB)^H = B^H A^H. This property is crucial for proving that the inverse of a Hermitian matrix is also Hermitian, as seen in the later steps of the solution.
Identity Matrix
An identity matrix, denoted as I, plays the role of 1 in matrix multiplication. Just as multiplying any number by 1 yields the same number, multiplying any square (n x n) matrix by the corresponding identity matrix (of the same dimension) will yield the original matrix. For every n-dimensional square matrix A, we have AI = IA = A.

The identity matrix contains 1s on the diagonal and 0s elsewhere. Due to its simplicity and unique properties, the identity matrix is a pivotal element in the discussion of matrix inverses. It is also invariant under conjugate transpose, which means I^H = I, a property used in the exercise to show that the inverse of a Hermitian matrix is itself Hermitian.

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 \(V\) be a finite dimensional vector space over the field \(K\), with a nondegenerate bilinear form \(\langle,\),\(rangle . If A: V \rightarrow V\) is a linear map such that $$ \langle A v, A w\rangle=\langle v, w\rangle $$ for all \(v, w \in V\), show that \(\operatorname{Det}(A)=\pm 1\).

Let \(A\) be a diagonal real unitary matrix. Show that the diagonal elements of \(A\) are equal to 1 or \(-1\).

A square \(n \times n\) real symmetric matrix \(A\) is said to be positive definite if \(^{\prime} X A X>0\) for all \(X \neq 0 .\) If \(A, B\) are symmetric (of the same size) we define \(A

(a) Let \(V\) be a finite dimensional space over \(\mathbf{R}\), with a positive definite scalar product. Let \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) and \(\left\\{w_{1}, \ldots, w_{n}\right\\}\) be orthonormal bases. Let \(A: V \rightarrow V\) be an operator of \(V\) such that \(A v_{i}=w_{i}\) Show that \(A\) is real unitary. (b) State and prove the analogous result in the complex case.

Let \(A, B\) be hermitian matrices (of the same size). Show that \(A+B\) is hermitian. If \(A B=B A\), show that \(A B\) is hermitian.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

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.