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Chapter 3: Problem 17

An \(n \times n\) matrix \(A\) has \(n\) eigenvalues \(A_{i} .\) If \(B=e^{A}\) show that \(B\) has the same eigenvectors as A with the corresponding eigenvalues \(B_{i}\) given by \(B_{i}=\exp \left(A_{i}\right)\). Note. \(e^{A}\) is defined by the Maclaurin expansion of the exponential: $$ e^{\mathrm{A}}=1+\mathrm{A}+\frac{\mathrm{A}^{2}}{2 !}+\frac{\mathrm{A}^{3}}{3 !}+\cdots $$

Short Answer

Expert verified
\( B \) has the same eigenvectors as \( A \), with eigenvalues \( B_i = e^{A_i} \).

Step by step solution

01

Review Definitions

Recall that if \( v \) is an eigenvector of matrix \( A \) with eigenvalue \( \lambda \), then \( Av = \lambda v \). We need to show that if \( B = e^A \), then \( v \) is also an eigenvector of \( B \) with eigenvalue \( e^\lambda \).
02

Write the Power Expansion for \( B = e^A \)

The Maclaurin series for \( e^A \) is given as \( e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots \), where \( I \) is the identity matrix.
03

Substitute Eigenvector Equation into the Series

Apply \( e^A \) to \( v \) using the series: \ \( e^A v = \left( I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots \right) v \ = v + Av + \frac{A^2v}{2!} + \frac{A^3v}{3!} + \cdots \).
04

Substitute Eigenvalue-Eigenvector Property

Since \( Av = \lambda v \), substitute \( \lambda \) for \( A \) multiplied by \( v \): \( e^A v = v + \lambda v + \frac{\lambda^2 v}{2!} + \frac{\lambda^3 v}{3!} + \cdots \).
05

Simplify Eigenvector Transformation

Factor out the common eigenvector \( v \), \( e^A v = \left( 1 + \lambda + \frac{\lambda^2}{2!} + \cdots \right) v \). This series simplifies to \( e^\lambda \), therefore \( e^A v = e^\lambda v \).
06

Conclusion on Eigenvectors and Eigenvalues

This shows that \( v \) is an eigenvector of \( B = e^A \) with the eigenvalue \( e^\lambda \). Hence, the eigenvectors of \( A \) and \( B \) are the same, while the eigenvalues \( \lambda \) of \( A \) correspond to \( e^\lambda \) for \( B \).

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Key Concepts

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

Matrix Exponential
The concept of a matrix exponential is crucial in solving systems of linear differential equations. Given a square matrix \(A\), we define the matrix exponential \(e^A\) using a power series expansion, analogous to the scalar exponential function. This is expressed as:\[ e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots \]where \(I\) is the identity matrix. This series continues infinitely and converges for any square matrix \(A\).
  • The matrix exponential shares properties with the standard exponential function, such as \(e^0 = I\) and \(e^{A+B} = e^A e^B\) if \(A\) and \(B\) commute.
  • It is particularly useful in finding solutions to linear systems of equations described by \( \frac{d}{dt}x(t) = Ax(t) \).
  • The concepts of eigenvalues and eigenvectors play a fundamental role in understanding how \(e^A\) affects a vector \(v\).
Understanding matrix exponentials helps bridge differential equations and linear algebra, providing a powerful tool for modeling dynamic systems in science and engineering.
Maclaurin Series
The Maclaurin series is a power series expansion used to express functions as an infinite sum of terms calculated from the values of the function's derivatives at a single point. For matrix exponentials, the Maclaurin series becomes a key tool.Given a function \(f(x)\), the Maclaurin series expansion is:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]
  • In the context of matrix exponentials, we apply this concept to derive \(e^A\), expanding \(A\) into powers just as with a scalar \(x\).
  • This series allows us to compute \(e^A\) by summing various powers of \(A\).
A deep understanding of Maclaurin series helps in computing matrix exponentials efficiently, especially when analytical solutions are needed for complex linear systems. By using this series, exponential operations involving matrices become more accessible.
Linear Algebra
Linear algebra is the mathematical framework for studying vectors, vector spaces, linear transformations, and systems of linear equations. It provides essential tools to understand eigenvalues and eigenvectors.
  • An eigenvector \(v\) of a matrix \(A\) is non-zero and satisfies the equation \(Av = \lambda v\), where \(\lambda\) is the eigenvalue.
  • Matrix exponentials \(e^A\) preserve eigenvectors of \(A\). They transform eigenvalues into their exponential counterparts, converting \(\lambda\) to \(e^\lambda\).
Linear algebra simplifies the analysis of linear transformations and systems represented by matrices. Understanding this topic is crucial to grasp how dynamic systems evolve over time and how transformations and scaling work in any vector space.
Power Series Expansion
A power series is an infinite sum of terms in the form \(a_n x^n\), acting as an essential tool in numerical and analytical analysis. Power series expansion enables us to approximate and analyze functions like \(e^x\), sine, and cosine effectively.
  • Power series turn complex functions into manageable polynomial forms, facilitating calculus operations such as differentiation and integration.
  • In matrix algebra, expanding \(e^A\) through a power series enables computations that go beyond regular scalar expressions.
  • This technique provides a method for deriving matrix decompositions, instrumental in approximating \(e^A\) and solving linear equations.
With power series expansion, mathematicians can tackle problems related to convergence and function approximation under varied conditions. This concept enhances our ability to utilize infinite sums for practical computations and theoretical insights.

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Most popular questions from this chapter

Write a subroutine that will form the adjoint of a complex \(M \times N\) matrix.

Unit masses are at the eight corners of a cube \((\pm 1, \pm 1, \pm 1)\). Find the moment of inertia matrix and show that there is a triple degeneracy. This means that so far as moments of inertia are concerned, the cubic structure exhibits spherical symmetry.

A matrix \(\mathrm{P}\) is a projection operator satisfying the condition $$ \mathrm{p}^{2}=\mathrm{P} $$ Show that the corresponding eigenvalues \(\left(\rho^{2}\right)_{2}\) and \(\rho_{2}\) satisfy the relation $$ \left(\rho^{2}\right)_{\lambda}=\left(\rho_{\lambda}\right)^{2}=\rho_{\lambda} $$ This means that the eigenvalues of \(P\) are 0 and 1 .

\(\mathrm{A}\) mass \(m_{1}=\frac{1}{2} \mathrm{~kg}\) is located at \((1,1,1)\) (meters), a mass \(m_{2}=\frac{1}{2} \mathrm{~kg}\) is at \((-1,-1,-1)\). The two masses are held together by an ideal (weightless, rigid) rod. (a) Find the inertia tensor of this pair of masses. (b) Find the eigenvalues and eigenvectors of this inertia matrix. (c) Explain the meaning, the physical significance of the \(\lambda=0\) eigenvalue. What is the significance of the corresponding eigenvector? (d) Now that you have solved this problem by rather sophisticated matrix techniques, explain how you could obtain (1) \(\lambda=0\) and \(\lambda=?\) by inspection (i.e., using common sense); and (2) \(\mathbf{r}_{\lambda=0}=\) ? by inspection (i.e., using freshman physics).

A and \(\mathrm{B}\) are two noncommuting Hermitian matrices: $$ A B-B A=i C $$ Prove that \(C\) is Hermitian.
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