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Chapter 9: Problem 9

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

Short Answer

Expert verified
Eigenvalues: \( \lambda = 1, 1.5 \pm 1.936i \).

Step by step solution

01

Define the Characteristic Polynomial

The characteristic polynomial of a matrix \( A \) is determined using the formula \( \text{det}(A - \lambda I) \), where \( \lambda \) is a scalar and \( I \) is the identity matrix of the same size as \( A \). For the matrix \( A = \begin{pmatrix} 1 & 2 & 3 \ 0 & 0 & 2 \ 0 & 3 & 1 \end{pmatrix} \), calculate \( A - \lambda I \):\[ A - \lambda I = \begin{pmatrix} 1-\lambda & 2 & 3 \ 0 & -\lambda & 2 \ 0 & 3 & 1-\lambda \end{pmatrix} \]
02

Compute the Determinant

To find the characteristic polynomial, compute the determinant of \( A - \lambda I \):\[ \text{det}(A - \lambda I) = \begin{vmatrix} 1-\lambda & 2 & 3 \ 0 & -\lambda & 2 \ 0 & 3 & 1-\lambda \end{vmatrix} \]Using the determinant formula for a 3x3 matrix:\[ = (1-\lambda) \left( (-\lambda)(1-\lambda) - 6 \right) \]\[ = (1-\lambda)(\lambda^2 - \lambda - 6) \]
03

Expand the Polynomial

Expand the polynomial obtained from the determinant calculation:\[ (1-\lambda)(\lambda^2 - \lambda - 6) = (1 - \lambda)(\lambda^2 - \lambda - 6) \]Distribute the term \((1-\lambda)\):\[ = (1)(\lambda^2 - \lambda - 6) - \lambda(\lambda^2 - \lambda - 6) \]\[ = \lambda^2 - \lambda - 6 - (\lambda^3 - \lambda^2 - 6\lambda) \]\[ = -\lambda^3 + 2\lambda^2 + 5\lambda - 6 \]This is the characteristic polynomial.
04

Find the Eigenvalues

To find the eigenvalues, solve the characteristic equation \(-\lambda^3 + 2\lambda^2 + 5\lambda - 6 = 0\). Factoring the polynomial can be complex, so consider testing rational roots and possible factorization. Try \( \lambda = 1 \):\[ -(1)^3 + 2(1)^2 + 5(1) - 6 = -1 + 2 + 5 - 6 = 0 \]Thus, \( \lambda = 1 \) is a root. Using synthetic division or polynomial division, factor the polynomial:\[ (\lambda - 1)(-\lambda^2 + 3\lambda - 6) = 0 \]Solve \(-\lambda^2 + 3\lambda - 6 = 0\) using the quadratic formula for the remaining roots:\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4(-1)(-6)}}{2(-1)} \]\[ \lambda = \frac{-3 \pm \sqrt{9 - 24}}{-2} \]\[ \lambda = \frac{-3 \pm \sqrt{-15}}{-2} \]Since the discriminant is negative, these roots are complex.
05

Summarize the Eigenvalues

The real eigenvalue is \( \lambda = 1 \), and the complex eigenvalues stem from the roots of \( -\lambda^2 + 3\lambda - 6 = 0 \), which are \( 1.5 \pm 1.936i \).

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Key Concepts

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

Eigenvalues
Eigenvalues are special numbers associated with a matrix. They represent the scalar values that satisfy the equation \( A\mathbf{v} = \lambda\mathbf{v} \), where \( A \) is the matrix, \( \lambda \) are the eigenvalues, and \( \mathbf{v} \) is a non-zero vector known as an eigenvector. In simpler terms, when you multiply a matrix by one of its eigenvectors, the vector does not change direction, and it's simply scaled by the eigenvalue.
  • They are crucial in understanding the properties of transformations described by the matrix.
  • Eigenvalues can tell us a lot about a matrix, such as if it can be inverted, or how it transforms objects in vector space.
To find the eigenvalues of a matrix, we typically solve the characteristic equation obtained from the characteristic polynomial. This involves computing the determinant of \( A - \lambda I \) and finding for which values of \( \lambda \) this determinant equals zero.
Determinant of a Matrix
The determinant of a matrix is a scalar value that can be calculated from its elements. It is a crucial value that provides important information about the matrix.
  • The determinant helps in finding the inverse of a matrix. If the determinant is zero, the matrix is non-invertible.
  • It plays a vital role in calculating the characteristic polynomial, which in turn is essential for determining eigenvalues.
For a 3x3 matrix like our example, the determinant is calculated using the rule for expansion along a row or a column. The formula involves multiplying the elements of the matrix by their corresponding minors and co-factors.In our problem, we take the 3x3 matrix \( A - \lambda I \) and calculate its determinant to form the characteristic polynomial. This polynomial is then solved to find the eigenvalues of the matrix.
Polynomial Factorization
Polynomial factorization is the process of breaking down a polynomial into simpler polynomials that multiply together to give the original polynomial. It is a vital step when solving the characteristic polynomial to find eigenvalues.
  • Factorization can sometimes be straightforward, especially if we have roots that are simple to determine.
  • Trial and error, synthetic division, and grouping are often used to factor polynomials when the roots are not immediately apparent.
In our example, once \( \lambda = 1 \) was found to be a root through substitution and checking, we used this information to factor the cubic polynomial partly. After factoring out \( \lambda - 1 \), synthetic division or direct division helps reduce the polynomial to further solve for other roots, either real or complex.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It is used when factorization or simple inspection doesn't readily solve the polynomial. The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
  • This formula provides both real and complex roots, depending on the value of the discriminant \( b^2 - 4ac \).
  • If the discriminant is positive, the roots are real and distinct; if zero, they are real and identical; and if negative, the roots are complex.
In the exercise, when encountering the quadratic \( -\lambda^2 + 3 \lambda - 6 = 0 \), we utilized the quadratic formula to determine the eigenvalues, resulting in complex roots due to the negative discriminant. This highlights the formula's usage in finding all possible solutions of a quadratic equation even when they are not visually obvious.

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Most popular questions from this chapter

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. \(A=\left(\begin{array}{rr}4 & 3 \\ -15 & -8\end{array}\right)\)

Each equation has a characteristic equation possessing distinct real roots. Find the general solution of each equation. \(y^{\prime \prime \prime}-2 y^{\prime \prime}-y^{\prime}+2 y=0\)

Do the following for each of the matrices in Exercises 26-33. Exercises \(26-29\) can be done by hand, but you should use a computer for the rest. (i) Find the eigenvalues. (ii) For each eigenvalue, find the algebraic and the geometric multiplicities. (iii) For each eigenvalue \(\lambda\), find the smallest integer \(k\) such that the dimension of the nullspace of \((A-\lambda I)^{k}\) is equal to the algebraic multiplicity. (iv) For each eigenvalue \(\lambda\), find \(q\) linearly independent generalized eigenvectors, where \(q\) is the algebraic multiplicity of \(\lambda\). (v) Verify that the collection of the generalized eigenvectors you find in part (iv) for all of the eigenvalues is linearly independent. (vi) Find a fundamental set of solutions for the system \(\mathbf{y}^{\prime}=\) Ay. \(A=\left(\begin{array}{rrr}-2 & 1 & -1 \\ 1 & -3 & 0 \\ 3 & -5 & 0\end{array}\right)\)

For each of the matrices, perform the following activities. (i) Determine where in the trace-determinate plane the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) fits. (ii) Find all of the equilibrium points for the system \(\mathbf{y}^{\prime}=A \mathbf{y}\), and plot them in the phase plane. (iii) Find the general solution and use the result to create a phase portrait for this system, taking care to sketch enough solution curves to show what happens in every part of the phase plane.\(A=\left(\begin{array}{rr}2 & 1 \\\ -10 & -5\end{array}\right)\)

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. \(\mathbf{y}^{\prime}=\left(\begin{array}{rr}6 & 4 \\ -1 & 2\end{array}\right) \mathbf{y}\)
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