/*! 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 76 Find real numbers \(x, y, z\) su... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 76

Find real numbers \(x, y, z\) such that \(A\) is Hermitian, where \(A=\left[\begin{array}{ccc}3 & x+2 i & y i \\ 3-2 i & 0 & 1+z i \\ y i & 1-x i & -1\end{array}\right]\)

Short Answer

Expert verified
The real numbers x, y, and z that make the matrix A Hermitian are \(x = 3\), \(y = 0\), and \(z = 3\).

Step by step solution

01

Compute the conjugate transpose of matrix A

First, we need to compute the conjugate transpose of matrix A, denoted by \(A^H\). To do this, first take the transpose of A, and then take the complex conjugate of each element of the matrix. Transpose of A: \(A^T=\left[\begin{array}{ccc}3 & 3-2 i & y i \\\ x+2 i & 0 & 1-x i \\\ y i & 1+z i & -1\end{array}\right]\) Complex conjugate of each element of \(A^T\): \(A^H=\left[\begin{array}{ccc}3 & 3+2 i & -y i \\\ x-2 i & 0 & 1+x i \\\ -y i & 1-z i & -1\end{array}\right]\)
02

Compare matrix A and its conjugate transpose

Now that we have the conjugate transpose of A, we can compare A and \(A^H\). For A to be Hermitian, its elements must equal the elements of \(A^H\). Hence: \(\left[\begin{array}{ccc}3 & x+2 i & y i \\\ 3-2 i & 0 & 1+z i \\\ y i & 1-x i & -1\end{array}\right] = \left[\begin{array}{ccc}3 & 3+2 i & -y i \\\ x-2 i & 0 & 1+x i\\\ -y i & 1-z i & -1\end{array}\right)\)
03

Extract the real numbers x, y, and z

By comparing the corresponding elements of A and \(A^H\), we can find the values of x, y, and z that make A Hermitian: 1. Comparing the (1, 2) element in A with the (1, 2) element in \(A^H\): \(x + 2i = 3 + 2i\). Therefore, \(x = 3\). 2. Comparing the (1, 3) element in A with the (1, 3) element in \(A^H\): \(yi = -yi\). This implies that \(y = 0\). 3. Comparing the (2, 3) element in A with the (2, 3) element in \(A^H\): \(1 + zi = 1 + xi\). Since we already know \(x = 3\), we have \(1 + zi = 1 + 3i\). Therefore, \(z = 3\). So the real numbers x, y, and z are 3, 0, and 3, respectively.

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.

Conjugate Transpose
The concept of the conjugate transpose is critical in understanding matrices involving complex numbers. When we talk about the conjugate transpose, we refer to two main operations: taking the transpose of the matrix and then finding the complex conjugate for each element. Here鈥檚 how you can do it:
  • Transpose: Flipping a matrix over its diagonal. For instance, if a matrix has a row of values, they become a column in the transposed version.
  • Complex Conjugate: For any element within the matrix that鈥檚 a complex number, you change the sign of the imaginary part. For example, if you have an element such as \(3 + 2i\), its complex conjugate will be \(3 - 2i\).

The resulting matrix from these operations is known as the conjugate transpose. If the original matrix is denoted as \(A\), the conjugate transpose is often represented as \(A^H\). This is a foundational step when working with Hermitian matrices, like in the exercise you have encountered, as Hermitian matrices are equal to their own conjugate transpose.
Complex Numbers
Complex numbers consist of two parts: a real part and an imaginary part. The general form for a complex number is \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part. Here are some key points to remember:
  • Real Part: This is the "ordinary" part of the complex number, like 3 in the example \(3 + 4i\).
  • Imaginary Part: This includes the imaginary unit \(i\), defined as \(\sqrt{-1}\), making operations beyond real arithmetic possible.'
  • Complex Conjugate: Important in matrix operations, it involves changing \(+bi\) to \(-bi\).

Complex numbers are essential in linear algebra when working with certain types of matrices. They help in defining and analyzing matrices with more nuanced properties, like Hermitian matrices. Understanding them is crucial for calculations involving the conjugate transpose and specific matrix comparisons.
Matrix Comparison
Comparing matrices involves checking if the corresponding elements in two matrices are equal. This step is essential in determining if a matrix meets certain criteria, such as being Hermitian. For matrix comparison:
  • Two matrices are said to be equal if they have the same dimensions and all corresponding elements are equal.
  • For Hermitian matrices, checking equality involves comparing the original matrix \(A\) with its conjugate transpose \(A^H\).
  • If \(A = A^H\) for all corresponding elements, then \(A\) is Hermitian. Simple comparisons of the form \(a+bi = c+di\) will involve equalizing real and imaginary parts separately.

In exercises involving Hermitian matrices, this kind of comparison is used to find unknowns, ensuring that they satisfy the unique properties of Hermitian matrices. For instance, comparing the elements lets us extract real numbers like \(x, y,\) and \(z\), ensuring \(A\) conforms to Hermitian properties.

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 all real triangular matrices \(A\) such that \(A^{2}=B,\) where (a) \(B=\left[\begin{array}{ll}4 & 21 \\ 0 & 25\end{array}\right],\) (b) \(B=\left[\begin{array}{rr}1 & 4 \\ 0 & -9\end{array}\right]\)

Let \(V\) and \(W\) be vector spaces with subspaces \(V_{1}\) and \(W_{1}\), respectively. If \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) is linear, prove that \(\mathrm{T}\left(\mathrm{V}_{1}\right)\) is a subspace of \(\mathrm{W}\) and that $\left\\{x \in \mathrm{V}: \mathrm{T}(x) \in \mathrm{W}_{1}\right\\}\( is a subspace of \)\mathrm{V}$.

Show that \(C^{\infty}\) is a subspace of \(\mathcal{F}(R, C)\).

Let \(A\) be invertible. Prove that \(A^{t}\) is invertible and \(\left(A^{t}\right)^{-1}=\left(A^{-1}\right)^{t}\). Visit goo.gl/suFm6V for a solution.

Pendular Motion. It is well known that the motion of a pendulum is approximated by the differential equation $$ \theta^{\prime \prime}+\frac{g}{l} \theta=0, $$where \(\theta(t)\) is the angle in radians that the pendulum makes with a vertical line at time \(t\) (see Figure 2.8), interpreted so that \(\theta\) is positive if the pendulum is to the right and negative if the pendulum is to the left of the vertical line as viewed by the reader. Here \(l\) is the length of the pendulum and \(g\) is the magnitude of acceleration due to gravity. The variable \(t\) and constants \(l\) and \(g\) must be in compatible units (e.g., \(t\) in seconds, \(l\) in meters, and \(g\) in meters per second per second). (a) Express an arbitrary solution to this equation as a linear combination of two real-valued solutions. (b) Find the unique solution to the equation that satisfies the conditions $$ \theta(0)=\theta_{0}>0 \text { and } \theta^{\prime}(0)=0 \text {. } $$ (The significance of these conditions is that at time \(t=0\) the pendulum is released from a position displaced from the vertical by \(\theta_{0}\).) (c) Prove that it takes \(2 \pi \sqrt{l / g}\) units of time for the pendulum to make one circuit back and forth. (This time is called the period of the pendulum.)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

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.