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Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1–8. $$ \left[\begin{array}{cc}{3} & {-2} \\ {1} & {-1}\end{array}\right] $$

Short Answer

Expert verified
The characteristic polynomial is \( \lambda^2 - 2\lambda - 1 \), and the eigenvalues are \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \).

Step by step solution

01

Define the Matrix

Consider the given matrix: \[ A = \begin{bmatrix} 3 & -2 \ 1 & -1 \end{bmatrix} \] The goal is to find its characteristic polynomial and the eigenvalues.
02

Set Up the Characteristic Equation

The characteristic polynomial is found by solving \[ \det(A - \lambda I) = 0 \] where \( I \) is the identity matrix of the same size as \( A \). For a 2x2 matrix, \[ A - \lambda I = \begin{bmatrix} 3 - \lambda & -2 \ 1 & -1 - \lambda \end{bmatrix} \] We'll determine the determinant of this matrix.
03

Calculate the Determinant

The determinant of \( A - \lambda I \) is calculated as:\[ \det(A - \lambda I) = (3 - \lambda)(-1 - \lambda) - (-2)(1) \] Evaluate the expression: \[ (3 - \lambda)(-1 - \lambda) = \lambda^2 - 2\lambda - 3 \] Adding the value from the second term: \[ \det(A - \lambda I) = \lambda^2 - 2\lambda - 3 + 2 = \lambda^2 - 2\lambda - 1 \] This is the characteristic polynomial.
04

Solve the Characteristic Polynomial for Eigenvalues

To find the eigenvalues, solve the equation:\[ \lambda^2 - 2\lambda - 1 = 0 \] Using the quadratic formula:\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -2, c = -1 \). Calculate:- \( b^2 - 4ac = (-2)^2 - 4(1)(-1) = 4 + 4 = 8 \)- \( \lambda = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \)The eigenvalues are \( \lambda_1 = 1 + \sqrt{2} \) and \( \lambda_2 = 1 - \sqrt{2} \).

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Key Concepts

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

Characteristic Polynomial
The characteristic polynomial is a crucial component in understanding the nature of a matrix. When dealing with matrices, especially when trying to find eigenvalues, the characteristic polynomial plays a significant role. It is essentially a polynomial obtained from the characteristic equation, \[ \det(A - \lambda I) = 0 \]where \( A \) is our matrix and \( \lambda \) represents the eigenvalues to be determined. To construct this polynomial, you subtract \( \lambda \) times the identity matrix \( I \) from your matrix \( A \), and then calculate the determinant of the resulting matrix. This determinant, set to zero, forms the characteristic polynomial.

This polynomial provides insights into various attributes of the matrix, such as stability, by revealing its eigenvalues. In our matrix, we derived the characteristic polynomial as: \[ \lambda^2 - 2\lambda - 1 \]. This expression will guide us in calculating the eigenvalues, which are the roots of the polynomial. Knowing how to find and manipulate characteristic polynomials is essential because it helps uncover fundamental properties of matrices and can be applied in numerous mathematical and real-world scenarios.
Determinant
The determinant is a scalar value that can be calculated from the elements of a square matrix. It provides specific important insights into the matrix, such as whether the matrix is invertible or not. For a 2x2 matrix like the one we have, the determinant is calculated using a simple formula:\[ \det \begin{bmatrix} a & b \ c & d \end{bmatrix} = ad - bc \]

Here, the determinant is used as part of finding the characteristic polynomial. In our case, the matrix we need to evaluate after substituting \( \lambda \) was:\[ \begin{bmatrix} 3 - \lambda & -2 \ 1 & -1 - \lambda \end{bmatrix} \]Then,\[ \det(A - \lambda I) = (3 - \lambda)(-1 - \lambda) - (-2)(1) \]calculating which gives us the expression \( \lambda^2 - 2\lambda - 1 \). The determinant in this context is fundamental to revealing the polynomial which, when solved, indicates the eigenvalues and thus critical characteristics of the matrix.
Matrix Equation
Matrix equations are mathematical expressions involving matrices that are solved to find unknowns, similar to solving regular algebraic equations. In the context of finding eigenvalues, the specific matrix equation at play is setting the determinant of \( A - \lambda I \) to zero, yielding the characteristic equation:\[ \det(A - \lambda I) = 0 \]

This equation forms the foundation of finding eigenvalues, as it provides a structured means to transform the matrix's properties into a solvable polynomial equation. Solving this relies on utilizing familiar algebraic techniques like the quadratic formula. For our example matrix, after forming \( \lambda^2 - 2\lambda - 1 = 0 \), we apply the quadratic formula:\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -2 \), and \( c = -1 \). Solving it gives eigenvalues \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \).

Understanding matrix equations helps decode complex systems and can be applied in various fields, such as quantum mechanics, economics, and data analysis.

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Most popular questions from this chapter

Another estimate can be made for an eigenvalue when an approximate eigenvector is available. Observe that if \(A \mathbf{x}=\lambda \mathbf{x}\) , then \(\mathbf{x}^{T} A \mathbf{x}=\mathbf{x}^{T}(\lambda \mathbf{x})=\lambda\left(\mathbf{x}^{T} \mathbf{x}\right),\) and the Rayleigh quotient $$R(\mathbf{x})=\frac{\mathbf{X}^{T} A \mathbf{x}}{\mathbf{x}^{T} \mathbf{x}}$$ equals \(\lambda\) . If \(\mathbf{x}\) is close to an eigenvector for \(\lambda,\) then this quotient is close to \(\lambda .\) When \(A\) is a symmetric matrix \(\left(A^{T}=A\right)\) , the Rayleigh quotient \(R\left(\mathbf{x}_{k}\right)=\left(\mathbf{x}_{k}^{T} A \mathbf{x}_{k}\right) /\left(\mathbf{x}_{k}^{T} \mathbf{x}_{k}\right)\) will have roughly twice as many digits of accuracy as the scaling factor \(\mu_{k}\) in the power method. Verify this increased accuracy in Exercises 11 and 12 by computing \(\mu_{k}\) and \(R\left(\mathbf{x}_{k}\right)\) for \(k=1, \ldots, 4\) $$ A=\left[\begin{array}{rr}{-3} & {2} \\ {2} & {0}\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l}{1} \\ {0}\end{array}\right] $$

A particle moving in a planar force field has a position vector \(x\) that satisfies \(\mathbf{x}^{\prime}=A \mathbf{x}\) . The \(2 \times 2\) matrix \(A\) has eigenvalues 4 and \(2,\) with corresponding eigenvectors \(\mathbf{v}_{1}=\left[\begin{array}{r}{-3} \\ {1}\end{array}\right]\) and \(\mathbf{v}_{2}=\left[\begin{array}{r}{-1} \\ {1}\end{array}\right] .\) Find the position of the particle at time \(t\) assuming that \(\mathbf{x}(0)=\left[\begin{array}{r}{-6} \\ {1}\end{array}\right]\)

In Exercises \(13-16,\) define \(T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) by \(T(\mathbf{x})=A \mathbf{x}\) . Find a basis \(\mathcal{B}\) for \(\mathbb{R}^{2}\) with the property that \([T]_{\mathcal{B}}\) is diagonal. $$ A=\left[\begin{array}{rr}{0} & {1} \\ {-3} & {4}\end{array}\right] $$

Suppose the eigenvalues of a \(3 \times 3\) matrix \(A\) are \(3,4 / 5,\) and \(3 / 5,\) with corresponding eigenvectors \(\left[\begin{array}{r}{1} \\ {0} \\\ {-3}\end{array}\right],\left[\begin{array}{r}{2} \\ {1} \\\ {-5}\end{array}\right],\) and \(\left[\begin{array}{r}{-3} \\ {-3} \\\ {7}\end{array}\right] .\) Let \(\mathbf{x}_{0}=\left[\begin{array}{r}{-2} \\\ {-5} \\ {3}\end{array}\right] .\) Find the solution of the equation \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) for the specified \(\mathbf{x}_{0},\) and describe what happens as \(k \rightarrow \infty\)

Classify the origin as an attractor, repeller, or saddle point of the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k} .\) Find the directions of greatest attraction and/or repulsion. \(A=\left[\begin{array}{rr}{1.7} & {-.3} \\ {-1.2} & {.8}\end{array}\right]\)

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