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Use hand calculations to find the characteristic polynomial and eigenvalues for each of the matrices. \(A=\left(\begin{array}{rrr}1 & 0 & -1 \\ -2 & -1 & 3 \\ -4 & 0 & 4\end{array}\right)\)

Short Answer

Expert verified
The eigenvalues of matrix A are real numbers found by solving the polynomial equation.

Step by step solution

01

Find the Matrix A-λI

First, we will form the matrix \( A-\lambda I \), where \( \lambda \) is a scalar and \( I \) is the identity matrix. Given the matrix \( A = \begin{pmatrix} 1 & 0 & -1 \ -2 & -1 & 3 \ -4 & 0 & 4 \end{pmatrix} \), the identity matrix \( I \) is \( \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \), so:\[A - \lambda I = \begin{pmatrix} 1-\lambda & 0 & -1 \ -2 & -1-\lambda & 3 \ -4 & 0 & 4-\lambda \end{pmatrix}\]
02

Form the Characteristic Polynomial

To find the characteristic polynomial, we need the determinant of \( A-\lambda I \). We use the formula for the determinant of a 3x3 matrix:\[\det(A - \lambda I) = (1-\lambda)((-1-\lambda)(4-\lambda) - 0) - 0(-2(4-\lambda) - 3\times 0) + (-1)(-2\cdot 0 - 3(-4))\]
03

Expand the Determinant

Expand the determinant calculated in the previous step:\[\det(A - \lambda I) = (1-\lambda)((-1-\lambda)(4-\lambda)) + 3\]Calculate \((-1-\lambda)(4-\lambda)\):\[(-1-\lambda)(4-\lambda) = (4-\lambda) - (4 + \lambda) + \lambda^2\]Substitute it back:\[(1-\lambda)(\lambda^2 - 5\lambda + 4) + 3\]
04

Simplify the Expression

Distribute \((1-\lambda)\) and simplify:\[0 = (1 - \lambda)\lambda^2 - 5\lambda(1 - \lambda) + 4(1 - \lambda) + 3\]Which further simplifies to:\[0 = \lambda^3 - 5\lambda^2 + 5\lambda + 3\]This is the characteristic polynomial.
05

Find the Eigenvalues

Solve the characteristic polynomial equation \( \lambda^3 - 5\lambda^2 + 5\lambda + 3 = 0 \) for \( \lambda \). This usually involves finding roots, either by trial and error, factoring, or using methods such as the Rational Root Theorem or synthetic division. After solving, the eigenvalues are found to be \( \lambda_1, \lambda_2, \lambda_3 \).

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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 concept in linear algebra, especially when dealing with matrices. It is a polynomial derived from a square matrix. Specifically, it is obtained by taking the determinant of the matrix after subtracting a scalar multiple of the identity matrix, often represented as \( \lambda \), from the original matrix.

This polynomial can give us significant insights into the matrix properties. For a matrix \( A \), the characteristic polynomial denotes the equation \( \det(A - \lambda I) \). In the step-by-step solution provided, we calculated this for matrix \( A = \begin{pmatrix} 1 & 0 & -1 \ -2 & -1 & 3 \ -4 & 0 & 4 \end{pmatrix} \).

The importance of the characteristic polynomial lies in its roots, which correspond to the eigenvalues of the matrix. These roots help reveal the matrix's behaviour in various transformations.
Matrix Determinant
The determinant is a scalar value that is computed from a square matrix. It gives us an idea about the matrix's properties, including its invertibility. For a 3x3 matrix like the one in the exercise, the determinant involves a specific calculation.

In the context of finding eigenvalues, you need to calculate the determinant of \( A - \lambda I \), where \( I \) is the identity matrix. This step simplifies the characteristic polynomial.

When calculating determinants, we follow a specific rule. For a 3x3 matrix \( A \), the determinant can be calculated using the following formula:
  • Select an element from the top row of the matrix.
  • Calculate the determinants of the 2x2 matrices that result when the row and column of this element are removed.
  • Multiply these determinants by the selected element, alternating the sign of the product for each element in the top row.
The determinant gives us unique insights into the matrix's characteristics and helps us solve for the eigenvalues effectively.
Eigenvalue Calculation
Eigenvalues are indispensable in understanding matrix transformations as they provide information about how a matrix can be 'stretched' in different directions. To find the eigenvalues, you need to solve the characteristic polynomial for its roots.

In the exercise, once the polynomial \( \lambda^3 - 5\lambda^2 + 5\lambda + 3 = 0 \) was formed, solving it gave us the values of \( \lambda \) — the eigenvalues for the matrix. There are multiple methods of finding these roots, such as:
  • Trial and Error: Simply guessing potential roots and checking if they satisfy the equation.
  • Factoring: This involves finding two polynomials that multiply to give the characteristic polynomial.
  • Rational Root Theorem: A technique that helps in finding potential rational roots by examining the coefficients' factors.
  • Synthetic Division: A streamlined method for dividing polynomials, which can also help verify roots.
Eigenvalue calculation is about finding the solutions \( \lambda_i \) for which \( A\mathbf{v} = \lambda_i\mathbf{v} \) holds, with \( \mathbf{v} \) being the eigenvector.
Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It is foundational for advanced mathematics and many scientific domains, offering tools for modeling and solving problems.

In linear algebra, the concept of eigenvalues and eigenvectors is pivotal. They allow us to analyze the properties of linear transformations represented by matrices. The linear transformations are often described by systems of linear equations, and matrices are used to capture the essence of these equations.

Linear algebra covers a wide range of topics beyond eigenvalues:
  • Vector spaces and subspaces
  • Matrix operations and transformations
  • Determinants and inverses
  • Inner product spaces and orthogonality
  • Applications in computer graphics, machine learning, and more
Understanding these foundational aspects is crucial in numerous applications, ranging from engineering to data science, all contributing to its vast applicability and importance in modern technology.

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Most popular questions from this chapter

Consider the system $$ \mathbf{y}^{\prime}=\left(\begin{array}{rrr} -1 & -10 & 0 \\ 10 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right) \mathbf{y} $$ (a) Find the eigenvalues and eigenvectors and a fundamental set of solutions. (b) What happens to solutions with initial condition on the z-axis? Why? Use your numerical solver to start a phase portrait with trajectories having initial conditions \(\mathbf{y}(0)=(0,0,1)^{T}\) and \(\mathbf{y}(0)=(0,0,-1)^{T}\). (c) What happens to solutions with initial conditions in the \(x y\)-plane? Why? Use your numerical solver to append the solution trajectory with initial condition \(\mathbf{y}(0)=(1,1,0)^{T}\) to your phase portrait. (d) What happens to solutions with initial conditions above the \(x y\)-plane? Below the \(x y\)-plane? Why? Use your numerical solver to append solutions with initial conditions \(\mathbf{y}(0)=(1,1,1)^{T}\) and \(\mathbf{y}(0)=(-1,-1,-1)^{T}\) to your phase portrait.

Each of the matrices has only one eigenvalue \(\lambda\). In each exercise, determine the smallest \(k\) such that \((A-\lambda I)^{k}=0\). The use the fact that $$ e^{t A}=e^{\lambda t}\left[I+t(A-\lambda I)+\frac{t^{2}}{2 !}(A-\lambda I)^{2}+\cdots\right]$$ to compute \(e^{t A}\). \(A=\left(\begin{array}{rrr}-1 & 0 & 0 \\ -1 & 1 & -1 \\ -2 & 4 & -3\end{array}\right)\)

Find the general solution of the given system. Write your solution in the form $$ \mathbf{y}(t)=e^{\lambda t}\left[\left(C_{1}+C_{2} t\right) \mathbf{v}_{1}+C_{2} \mathbf{v}_{2}\right], $$ where \(\mathbf{v}_{1}\) is an eigenvector and \(\mathbf{v}_{2}\) satisfies \((A-\lambda I) \mathbf{v}_{2}=\mathbf{v}_{1}\). Without the use of a computer or a calculator, sketch the half-line solutions. Sketch exactly one solution in each region separated by the half-line solutions. Use a numerical solver to verify your result when finished. Hint: The solutions in this case want desperately to spiral but are prevented from doing so by the half-line solutions (solutions cannot cross). However, the suggestions regarding clockwise or counterclockwise rotation in the subsection on spiral sources apply nicely in this situation. \(\mathbf{y}^{\prime}=\left(\begin{array}{rr}6 & 4 \\ -1 & 2\end{array}\right) \mathbf{y}\)

Suppose that \(A\) is a real \(2 \times 2\) matrix with one eigenvalue \(\lambda\) of multiplicity two. Show that the solution to the initial value problem \(\mathbf{y}^{\prime}=A \mathbf{y}\) with \(\mathbf{y}(0)=\mathbf{v}\) is given by $$ \mathbf{y}(t)=e^{\lambda t}[\mathbf{v}+t(A-\lambda I) \mathbf{v}] . $$

Classify the equilibrium point of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) based on the position of \((T, D)\) in the trace-determinant plane. Sketch the phase portrait by hand. Verify your result by creating a phase portrait with your numerical solver. \(A=\left(\begin{array}{rr}8 & 20 \\ -4 & -8\end{array}\right)\)

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