/*! 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 5 Let \(\Psi=\) \(=\left[\begin{ar... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 5

Let \(\Psi=\) \(=\left[\begin{array}{ccc|c} & A & & 1 \\ & & & 1 \\ \hline 0 & \cdots & 0 & 1\end{array}\right]\). Prove that the eigenvalues of \(\Psi\) consist of the eigenvalues of \(A\) and 1 .

Short Answer

Expert verified
To prove that the eigenvalues of \(\Psi\) consist of the eigenvalues of A and 1, we first find the characteristic equation for \(\Psi\) by computing the determinant of \((\Psi - \lambda \text{I})\), where \(\lambda\) is the eigenvalue and I is the identity matrix. This gives us: \[\det(\Psi - \lambda \text{I}) = (1-\lambda) \cdot \det(A - \lambda \text{I})\] The determinant is zero when either \((1-\lambda) = 0\) or \(\det(A - \lambda \text{I}) = 0\). This leads to \(\lambda = 1\) and the eigenvalues of A, respectively. Therefore, the eigenvalues of \(\Psi\) include the eigenvalues of A and 1.

Step by step solution

01

Find the characteristic equation for \(\Psi\)

To find the characteristic equation for \(\Psi\), we need to compute the determinant of (\(\Psi\) - \(\lambda\)I), where \(\lambda\) is the eigenvalue and I is the identity matrix of the same size as \(\Psi\). This determinant will be given by: \[\det(\Psi - \lambda I) = \left|\begin{array}{ccc|c} A - \lambda I & 1 \\\ \hline 0 & \cdots & 0 & 1-\lambda\end{array}\right|\]
02

Simplify the determinant

Now we will simplify the determinant by focusing on the last row. By expanding along the last row, we get: \[\det(\Psi - \lambda I) = (1-\lambda) \cdot \det(A - \lambda I)\]
03

Solve for eigenvalues

Now, we want to find the eigenvalues by setting the determinant equal to 0: \[(1-\lambda) \cdot \det(A - \lambda I) = 0\] This equation holds when either \((1-\lambda) = 0\) or \(\det(A - \lambda I) = 0\). For \((1-\lambda) = 0\), we have only one solution, \(\lambda = 1\). For \(\det(A - \lambda I) = 0\), we have the same equation as the characteristic equation for A, which means the eigenvalues will be the eigenvalues of A. In conclusion, the eigenvalues of \(\Psi\) consist of the eigenvalues of A and 1.

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.

Characteristic Equation
Understanding the characteristic equation is crucial when learning about eigenvalues of matrices. For any square matrix, say a 2x2 or 3x3 matrix, the characteristic equation is formed by subtracting a scalar, denoted as \( \lambda \) (lambda), times the identity matrix from the original matrix, and setting the determinant of this difference to zero.

Formally, given a matrix \( A \), the characteristic equation is \( \det(A - \lambda I) = 0 \), where \( I \) represents the identity matrix of the same dimensions as matrix \( A \). This equation is pivotal because solving it provides us with the eigenvalues of \( A \), which are values of \( \lambda \) that make the equation true. In essence, these eigenvalues are the factors that, when \( A \) is scaled and transformed, maintain its direction, albeit stretching or compressing it.
Determinant Calculation
Determinant calculation is a process that associates a scalar value with a square matrix. In the context of finding eigenvalues, the determinant helps determine whether a vector, when transformed by the matrix, will change scale. For example, if the determinant is zero, the matrix will have at least one vector that becomes zero, indicating a loss of dimension.

The actual computation of the determinant can vary in complexity depending on the size of the matrix. For a 2x2 matrix, the determinant is quite straightforward to calculate. However, as matrices get larger, techniques like expansion by minors or row reduction can be employed to simplify the process. The determinant's role in finding eigenvalues is central to understanding the behavior of linear transformations represented by the matrix.
Eigenvalue Computation
After setting up the characteristic equation, the next step is the computation of eigenvalues. This involves finding the roots of the polynomial that the characteristic equation represents.

In the exercise provided, the computation simplified into two parts due to the block matrix structure of \( \Psi \). The problem bifurcated into finding \( \lambda = 1 \) and solving \( \det(A - \lambda I) = 0 \). For students, it's important to realize that each eigenvalue has an associated eigenvector, and these pairs fundamentally capture the essence of how the matrix transforms space.
Identity Matrix
The identity matrix, often denoted as \( I \), is a special square matrix with ones on the diagonal and zeros elsewhere. Its dimensions match the matrix it's associated with when determining eigenvalues.

The role of the identity matrix in finding eigenvalues cannot be overstated. It acts as the multiplicative identity in the matrix world, analogous to the number 1 in real number multiplication. When you subtract \( \lambda I \) from the matrix \( A \), the goal is to discern how much the matrix deviates from this basic form and thus identify the eigenvalues, which are the scalar values that describe this deviation.

Key Importance of Identity Matrix

The identity matrix is the key to unlocking the eigenvalues of a matrix. It enables the formation of the characteristic equation and the determination of the matrix's most fundamental 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

(For those who've thought about convergence issues) Check that the power series expansion $$ f(x)=\sum_{k=0}^{\infty} \frac{x^{k}}{k !} $$ converges for any real number \(x\) and that \(f(x)=e^{x}\), as follows. a. Fix \(x \neq 0\) and choose an integer \(K\) so that \(K \geq 2|x|\). Then show that for \(k>K\), we have \(\frac{|x|^{k}}{k !} \leq C\left(\frac{1}{2}\right)^{k-K}\), where \(C=\frac{|x|}{K} \cdot \ldots \cdot \frac{|x|}{2} \cdot \frac{|x|}{1}\) is a fixed constant. b. Conclude that the series \(\sum_{k=K+1}^{\infty} \frac{|x|^{k}}{k !}\) is bounded by the convergent geometric series \(C \sum_{j=1}^{\infty} \frac{1}{2}\) and therefore converges and, thus, that the entire original series converges absolutely. c. It is a fact that every convergent power series may be differentiated (on its interval of convergence) term by term to obtain the power series of the derivative (see Spivak, Calculus, Chapter 24). Check that \(f^{\prime}(x)=f(x)\) and deduce that \(f(x)=e^{x}\).

Prove that \(\operatorname{det}\left(e^{A}\right)=e^{\operatorname{tr} A}\). (Hint: First assume \(A\) is diagonalizable. In the general case, apply the result of Exercise \(6.2 .15\), which also works with complex matrices.)

In each case, give the \(3 \times 3\) matrix representing the isometry of \(\mathbb{R}^{2}\) and then use your answer to fit the isometry into our classification scheme. "a. First translate by \(\left[\begin{array}{l}1 \\ 1\end{array}\right]\), and then rotate \(\pi / 2\) about the point \((-1,0)\). "b. First reflect across the line \(x_{1}+x_{2}=1\), and then translate by \(\left[\begin{array}{l}1 \\ 1\end{array}\right]\). c. First rotate \(\pi / 4\) about the origin, and then reflect across the line \(x_{1}+x_{2}=1\).

Let \(A\) be a square matrix. a. Prove that \(A e^{t A}=e^{t A} A\). b. Prove that \(\left(e^{A}\right)^{-1}=e^{-A}\). (Hint: Differentiate the product \(e^{t A} e^{-t A}\).) c. Prove that if \(A\) is skew-symmetric (i.e., \(A^{\top}=-A\) ), then \(e^{A}\) is an orthogonal matrix.

Check that if \(A\) is an \(n \times n\) matrix and the \(n \times n\) differentiable matrix function \(E(t)\) satisfies \(\frac{d E}{d t}=A E(t)\) and \(E(0)=I\), then \(E(t)=e^{t A}\) for all \(t \in \mathbb{R}\).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

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.