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Chapter 4: Problem 2

Show that \(A\) and \(B\) are not similar matrices. \(A=\left[\begin{array}{rr}2 & 1 \\ -4 & 6\end{array}\right], B=\left[\begin{array}{rr}3 & -1 \\ -5 & 7\end{array}\right]\)

Short Answer

Expert verified
A and B are not similar because they have different eigenvalues.

Step by step solution

01

Verify Similarity Condition

Two matrices are similar if there exists an invertible matrix \(P\) such that \(P^{-1}AP = B\). However, it is more straightforward to check if they have the same eigenvalues since similar matrices have the same eigenvalues.
02

Find Eigenvalues of Matrix A

To find the eigenvalues of matrix \(A\), we calculate the characteristic polynomial. The characteristic equation is \(\det(A - \lambda I) = 0\). For \(A\), this becomes:\[\det\left(\begin{array}{cc} 2-\lambda & 1 \ -4 & 6-\lambda \end{array}\right) = 0\]This simplifies to \[(2-\lambda)(6-\lambda) - (1)(-4) = 0\,\lambda^2 - 8\lambda + 16 = 0.\]
03

Solve for Eigenvalues of Matrix A

Solve the quadratic equation \(\lambda^2 - 8\lambda + 16 = 0\). Using the quadratic formula, \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find:\[\lambda = \frac{8 \pm \sqrt{8^2 - 4 \times 1 \times 16}}{2} = \frac{8 \pm \sqrt{0}}{2} = 4.\]The eigenvalues of \(A\) are both 4.
04

Find Eigenvalues of Matrix B

Similarly, find the eigenvalues for matrix \(B\) by solving:\[\det\left(\begin{array}{cc} 3-\lambda & -1 \ -5 & 7-\lambda \end{array}\right) = 0.\]Expand this to get:\((3-\lambda)(7-\lambda) - (-1)(-5) = 0\),giving:\((3-\lambda)(7-\lambda) - 5 = 0\).This simplifies to:\[\lambda^2 - 10\lambda + 16 = 0.\]
05

Solve for Eigenvalues of Matrix B

Solve the equation \(\lambda^2 - 10\lambda + 16 = 0\) using the quadratic formula:\[\lambda = \frac{10 \pm \sqrt{10^2 - 4 \times 1 \times 16}}{2} = \frac{10 \pm \sqrt{52}}{2} = \frac{10 \pm 2\sqrt{13}}{2}.\]The eigenvalues of \(B\) are \(5 + \sqrt{13}\) and \(5 - \sqrt{13}\).
06

Compare Eigenvalues of A and B

Since the eigenvalues of \(A\) are both 4 and the eigenvalues of \(B\) are \(5 + \sqrt{13}\) and \(5 - \sqrt{13}\), the eigenvalues are not the same. Therefore, \(A\) and \(B\) are not similar because they do not share the same eigenvalues.

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Key Concepts

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

Eigenvalues
Eigenvalues are essential in understanding how a matrix can be transformed through linear algebra. For a matrix, eigenvalues are special scalars that emerge from solutions to the characteristic equation. To determine these values for a given matrix, say matrix \(A\), we solve the characteristic equation \(\det(A - \lambda I) = 0\). Here, \(\lambda\) denotes the eigenvalue in question, while \(I\) is the identity matrix of the same order as \(A\).
  • Eigenvalues can give insights into the stability and certain properties of a matrix.
  • In the context of matrix similarity, two matrices are only similar if they have the same eigenvalues.
Understanding and calculating eigenvalues are vital for determining various properties and behaviors associated with matrices.
Invertible Matrix
An invertible matrix is a square matrix that has an inverse, meaning there exists another matrix that can reverse its effect when multiplied together. The concept of invertibility is crucial because it determines whether a particular matrix transformation can be undone. For a matrix \(P\), the inverse \(P^{-1}\) exists only if \(P\) is non-singular or invertible.
  • A non-invertible matrix, also known as a singular matrix, has a determinant equal to zero.
  • Invertibility is a key component of matrix similarity, particularly in expressing that if two matrices \(A\) and \(B\) are similar, then \(P^{-1}AP = B\) for some invertible matrix \(P\).
The properties that ensure invertibility are significant when translating complex matrix operations or transformations into simpler, manipulatable forms.
Characteristic Equation
The characteristic equation of a matrix plays a fundamental role in determining its eigenvalues. Specifically, for a given matrix \(A\), the characteristic equation is formed by the determinant of the matrix \(A - \lambda I\), set equal to zero: \(\det(A - \lambda I) = 0\). Solving this equation results in the eigenvalues of the matrix.
  • This equation is often a polynomial in \(\lambda\), and its degree corresponds to the matrix's order.
  • The roots of the characteristic equation are the eigenvalues, which provide insights into the matrix's behavior.
Understanding how to form and solve the characteristic equation is crucial for analyzing the inherent properties of matrices.
Quadratic Formula
The quadratic formula is a significant mathematical tool used to find the roots of quadratic equations, which are polynomial equations of degree two. This formula is expressed as \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). It provides a straightforward way to solve equations of the form \(ax^2 + bx + c = 0\), which often arises in the computation of eigenvalues.
  • The quadratic formula can be used when the characteristic equation of a matrix takes on a quadratic form, often seen in 2x2 matrices.
  • It allows us to find the eigenvalues of matrices effectively, even when they result in complex numbers.
The application of the quadratic formula simplifies the process of finding eigenvalues and is indispensable in matrix algebra.

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Most popular questions from this chapter

Verify the Cayley-Hamilton Theorem for $$A=\left[\begin{array}{lll} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}\right]$$ The Cayley-Hamilton Theorem can be used to calculate powers and inverses of matrices. For example, if \(A\) is a \(2 \times 2\) matrix with characteristic polynomial \(c_{A}(\lambda)=\lambda^{2}+a \lambda+b\) \(\operatorname{then} A^{2}+a A+b I=0,\) so $$A^{2}=-a A-b I$$ and $$\begin{aligned} A^{3} &=A A^{2}=A(-a A-b I) \\ &=-a A^{2}-b A \\ &=-a(-a A-b I)-b A \\ &=\left(a^{2}-b\right) A+a b I \end{aligned}$$ It is easy to see that by continuing in this fashion we can express any positive power of \(A\) as a linear combination of I and A. From \(A^{2}+a A+b I=O,\) we also obtain \(A(A+a I)=-b I, s o\) $$A^{-1}=-\frac{1}{b} A-\frac{a}{b} I$$ provided \(b \neq 0\).

Consider the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\). (a) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\) (b) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 0\end{array}\right]\) (c) Using eigenvalues and eigenvectors, classify the origin as an attractor, repeller, saddle point, or none of these. (d) Sketch several typical trajectories of the system. $$A=\left[\begin{array}{rr} 0.2 & 0.4 \\ -0.2 & 0.8 \end{array}\right]$$

Write out the first six terms of the sequence defined by the recurrence relation with the given initial conditions. $$a_{1}=128, a_{n}=a_{n-1} / 2 \text { for } n \geq 2$$

Find the general solution to the given system of differential equations. Then find the specific solution that satisfies the initial conditions. (Consider all functions to be functions of t.) $$\begin{array}{l} x^{\prime}=x+\quad 3 z, \quad x(0)=2 \\ y^{\prime}=x-2 y+z, \quad y(0)=3 \\ z^{\prime}=3 x+\quad z, \quad z(0)=4 \end{array}$$

Let \(A\) be a nilpotent matrix (that is, \(A^{m}=O\) for some \(m>1\) ). Prove that if \(A\) is diagonalizable, then \(A\) must be the zero matrix.
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