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Chapter 5: Problem 99

Given that \(\lambda=1\) is a three-times repeated eigenvalue of the matrix $$ \boldsymbol{A}=\left[\begin{array}{rrr} -3 & -7 & -5 \\ 2 & 4 & 3 \\ 1 & 2 & 2 \end{array}\right] $$ determine how many independent eigenvectors correspond to this value of \(\lambda\). Determine a corresponding set of independent eigenvectors.

Short Answer

Expert verified
The eigenvalue \( \lambda = 1 \) has 1 independent eigenvector: \( \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \).

Step by step solution

01

Define the Eigenvalue Equation

To find the eigenvectors corresponding to an eigenvalue \( \lambda \), we start with the equation \( (\boldsymbol{A} - \lambda \boldsymbol{I}) \boldsymbol{x} = \boldsymbol{0} \). Here, \( \boldsymbol{I} \) is the identity matrix and \( \lambda = 1 \). Substitute \( \lambda \) and \( \boldsymbol{A} \) into the equation.
02

Setup the Matrix Equation

The matrix equation becomes \( (\boldsymbol{A} - \boldsymbol{I}) \boldsymbol{x} = \boldsymbol{0} \), or \[\begin{bmatrix}-4 & -7 & -5 \2 & 3 & 3 \1 & 2 & 1\end{bmatrix}\begin{bmatrix}x_1 \x_2 \x_3\end{bmatrix} = \begin{bmatrix}0 \0 \0\end{bmatrix}.\]
03

Reduce the Matrix to Row-Echelon Form

Perform Gaussian elimination on the matrix \(\begin{bmatrix}-4 & -7 & -5 \2 & 3 & 3 \1 & 2 & 1\end{bmatrix}\) to obtain its row-echelon form. This process helps identify the number of linearly independent solutions (eigenvectors).
04

Find the Reduced Row Form

After performing Gaussian elimination, we obtain:\[\begin{bmatrix}1 & 2 & 1 \0 & 1 & 1 \0 & 0 & 0\end{bmatrix}.\]This indicates a rank of 2, meaning two leading variables exist. Given our original 3x3 matrix, there are \(3 - 2 = 1\) free variables, thus the solution space is 1-dimensional.
05

Express Dependent Variables in Terms of Free Variables

Let \( x_3 = t \), where \( t \) is any real number. Then \( x_2 = -t \) and \( x_1 = t \/ 2 \). Therefore, a general solution (eigenvector) could be expressed in terms of \( t \): \[\begin{bmatrix}x_1 \x_2 \x_3\end{bmatrix} = t \begin{bmatrix}1 \-1 \1\end{bmatrix}\]
06

Determine a Set of Independent Eigenvectors

Since the solution is a scalar multiple of the vector \( \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} \), there is only one linearly independent eigenvector corresponding to \( \lambda = 1 \). Therefore, the vector \( \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} \) represents the eigenvector for \( \lambda = 1 \).
07

Confirm the Algebraic and Geometric Multiplicity

The eigenvalue \( \lambda = 1 \) has an algebraic multiplicity of 3 (it repeats three times). However, the geometric multiplicity, which is the number of linearly independent eigenvectors, is 1, as determined by the row-reduced matrix form.

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Key Concepts

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

Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors, vector spaces, linear transformations, and systems of linear equations. It's foundational for understanding various mathematical concepts and has applications in data science, engineering, physics, and more. In the context of eigenvalues and eigenvectors, linear algebra provides the tools to analyze matrices and their properties.
  • An **eigenvalue** is a scalar that indicates how much a vector is stretched or shrunk during a linear transformation represented by a matrix.
  • An **eigenvector** is a non-zero vector that changes at most by a scalar factor when that linear transformation is applied.
Understanding how to work with eigenvalues and eigenvectors is crucial for solving various systems of linear equations and understanding the structure of linear transformations.
Matrix Row Reduction
Matrix row reduction is a systematic method used to simplify matrices. It involves performing row operations to transform a given matrix into a simpler form, such as row-echelon form or reduced row-echelon form. This process is crucial for solving systems of linear equations, including finding eigenvectors.
To reduce a matrix,
  • Start by selecting a pivot element (usually the leftmost non-zero element in a row) to eliminate other non-zero elements in the same column.
  • Perform row operations such as swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting a multiple of one row to another.
The goal is to create zeros below each pivot position and transform the matrix into an upper triangular matrix. Once this form is achieved, we can determine the number of linearly independent vectors, as seen in the provided step-by-step solution.
Algebraic and Geometric Multiplicity
Algebraic and geometric multiplicity are key concepts when analyzing eigenvalues. They help determine the nature of solutions related to a given eigenvalue.
  • **Algebraic multiplicity** refers to the number of times a given eigenvalue appears as a root of the characteristic polynomial of a matrix. In this exercise, the eigenvalue \( \lambda = 1 \) has an algebraic multiplicity of 3, meaning it is repeated three times as a solution to the characteristic polynomial of matrix \( \boldsymbol{A} \).
  • **Geometric multiplicity** is the number of linearly independent eigenvectors associated with an eigenvalue. For \( \lambda = 1 \), after performing matrix reductions, we find the geometric multiplicity is 1, indicating the presence of only one independent eigenvector. This eigenvector can be expressed as \( \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} \).
Understanding these multiplicities is crucial, as they provide insights into the dimensions of eigenspaces and how transformations affect vector spaces. The practical implication here is that even if an eigenvalue repeats, it doesn't necessarily result in more independent eigenvectors.

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Most popular questions from this chapter

Obtain the characteristic polynomials of the matrices (a) \(\left[\begin{array}{rr}2 & -1 \\ -1 & 2\end{array}\right]\) (b) \(\left[\begin{array}{ll}2 & 1 \\ 1 & 1\end{array}\right]\) (c) \(\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 0 & 3\end{array}\right]\) (d) \(\left[\begin{array}{lll}1 & 2 & 0 \\ 0 & 2 & 2 \\ 0 & 1 & 3\end{array}\right]\) (e) \(\left[\begin{array}{lll}3 & 2 & 1 \\ 4 & 5 & -1 \\ 2 & 3 & 4\end{array}\right]\) (f) \(\left[\begin{array}{rr}2 & 1 \\ -1 & a\end{array}\right]\) and hence evaluate the eigenvalues of the matrices.

Express the determinant $$ \left|\begin{array}{ccc} \alpha & \beta & \gamma \\ \beta \gamma & \gamma \alpha & \alpha \beta \\ -\alpha+\beta+\gamma & \alpha-\beta+\gamma & \alpha+\beta-\gamma \end{array}\right| $$ as a product of linear factors.

Find the rank of the matrices (a) \(\left[\begin{array}{cccc}2 & 1 & 1 & 1 \\ 4 & 2 & 2 & 3 \\ 0 & 0 & 0 & 1 \\ -2 & -1 & -1 & 0\end{array}\right], \quad\) (b) \(\left[\begin{array}{cccc}1 & 1 & 1 & 1 \\ 2 & 1 & 2 & 1 \\ 0 & 1 & 0 & 1 \\\ 1 & 0 & 1 & 1\end{array}\right]\)

If $$ \begin{aligned} &\boldsymbol{A}=\left[\begin{array}{lll} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right], \quad \boldsymbol{B}=\left[\begin{array}{lll} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 5 \end{array}\right] \\ &\text { and } \boldsymbol{C}=\left[\begin{array}{rrr} 4 & 0 & -2 \\ 5 & 3 & 1 \\ 2 & 5 & 4 \end{array}\right] \end{aligned} $$ (a) show that $$ \operatorname{trace}(\boldsymbol{A}+\boldsymbol{B})=\operatorname{trace} \boldsymbol{A}+\operatorname{trace} \boldsymbol{B} $$ (b) find \(\boldsymbol{D}\) so that \(\boldsymbol{A}+\boldsymbol{D}=\boldsymbol{C}\) (c) verify the associative law $$ (\boldsymbol{A}+\boldsymbol{B})+\boldsymbol{C}=\boldsymbol{A}+(\boldsymbol{B}+\boldsymbol{C}) $$

Given the matrices $$ \boldsymbol{A}=\left[\begin{array}{lll} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 1 & 2 \end{array}\right] \text { and } \quad \boldsymbol{B}=\left[\begin{array}{rrr} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{array}\right] $$ evaluate \(\boldsymbol{C}\) in the three cases. (a) \(\boldsymbol{C}=\boldsymbol{A}+\boldsymbol{B}\) (b) \(2 A+3 C=4 B\) (c) \(A-C=B+C\)
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