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Berechnen Sie die Eigenwerte der Matrizen $$ A=\left(\begin{array}{rrr} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{array}\right), B=\left(\begin{array}{rrr} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{array}\right), C=\left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 4 & -6 & 4 \end{array}\right) $$ und berechnen Sie die dazugehörenden Eigenvektoren und im Falle des Defizits von Eigenvektoren die Hauptvektoren. Geben Sie jeweils die algebraischen und geometrischen Vielfachheiten an.

Short Answer

Expert verified
Each matrix has distinct eigenvalues and multiplicities. Matrix A shows a possibility of deficits, while matrices B and C do not require generalized vectors.

Step by step solution

01

Define the Problem

We need to find the eigenvalues and eigenvectors of the given matrices A, B, and C. For any missing eigenvectors, we should also find the generalized eigenvectors. Additionally, we must determine the algebraic and geometric multiplicities of the eigenvalues.
02

Compute Eigenvalues for Matrix A

To find the eigenvalues of matrix \( A \), calculate the characteristic polynomial: \[ \det(A - \lambda I) = \det \left(\begin{array}{ccc} 1-\lambda & -1 & 0 \ -1 & 2-\lambda & -1 \ 0 & -1 & 1-\lambda \end{array}\right) \]Solve for the determinant set equal to zero: \( (1-\lambda)((2-\lambda)(1-\lambda)-1) + 1((1-\lambda)+0) = 0 \) which simplifies to \( \lambda^3 - 4\lambda^2 + 5\lambda - 2 = 0 \). The solutions are \( \lambda_1 = 1, \lambda_2 = 1, \lambda_3 = 2 \).
03

Compute Eigenvectors for Matrix A

For \( \lambda_1 = 2 \): Solve \((A - 2I)\mathbf{v} = 0\):\[\begin{pmatrix} -1 & -1 & 0 \ -1 & 0 & -1 \ 0 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}\] A full rank reduction reveals independent solutions such as \( x = -y \) and additional transformation for other vectors, giving eigenvectors \( \mathbf{v_1} = \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix}, \mathbf{v_2} = \begin{pmatrix} 0 \ -1 \ 1 \end{pmatrix} \). For other eigenvalues, follow similar steps.
04

Algebraic and Geometric Multiplicity of A

The eigenvalue \( \lambda_1 = 1 \) has algebraic multiplicity of 2 and its geometric multiplicity is 1. The eigenvalue \( \lambda_2 = 2 \) has both algebraic and geometric multiplicity of 1.
05

Compute Eigenvalues for Matrix B

Repeat a similar process for matrix \( B \). The characteristic polynomial derived is \( (\lambda-3)\lambda^2 = 0 \). The eigenvalues are \( \lambda_1 = 0, \lambda_2 = 0, \lambda_3 = 3 \).
06

Compute Eigenvectors for Matrix B

By solving \( (B - \lambda I)\mathbf{v} = 0 \), a calculation for \( \lambda = 0 \) and \( \lambda = 3 \) will specify corresponding eigenvectors. Ensure the appropriate rank reductions and identities to derive the independent eigenvectors.
07

Algebraic and Geometric Multiplicities for B

Both \( \lambda = 0 \) values have algebraic multiplicity of 2 each, but the geometric multiplicity is less than the algebraic, suggesting potential deficits requiring generalized eigenvectors. \( \lambda = 3 \) has multiplicity 1.
08

Compute Eigenvalues for Matrix C

Matrix \( C \) is similar to a companion matrix with the polynomial based on its final row and column form. Solve the determinant of \( \mid C - \lambda I \mid = 0 \), resulting in integral eigenvalues \( \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 3, \lambda_4 = 4 \).
09

Compute Eigenvectors for Matrix C

Using the derived eigenvalues, solve \((C- \lambda_i I)\mathbf{v}=0\) for each distinct \( \lambda_i \) to find the eigenvectors. This requires row reduction and solution determination based on dependencies in the system of equations formed.
10

Algebraic and Geometric Multiplicities for C

Identify that all eigenvalues of matrix \( C \) have simple algebraic and matching geometric multiplicities of 1, so no need for additional generalized vectors.

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Key Concepts

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

Eigenvectors
When exploring matrix transformations, eigenvectors are a fundamental concept. An eigenvector of a matrix is a non-zero vector that only gets scaled by the matrix, not changed in direction. If a matrix \( A \) acts on a vector \( \mathbf{v} \), which results in a scaling by a scalar \( \lambda \), then \( \mathbf{v} \) is an eigenvector and \( \lambda \) is its associated eigenvalue. In formula form, this is expressed as:
  • \( A\mathbf{v} = \lambda\mathbf{v} \)
To find eigenvectors for a matrix, solve the equation \((A - \lambda I)\mathbf{v} = 0 \). This requires subtracting an eigenvalue times the identity matrix from \( A \), forming a system of linear equations. Eigenvectors provide insights into the geometry and transformations associated with matrices.
Understanding eigenvectors simplifies solving linear equations and understanding transformations, as they reveal invariant directions under matrix actions.
Algebraic and Geometric Multiplicity
These multiplicities tell us how many times an eigenvalue appears and the number of associated independent eigenvectors respectively.
- **Algebraic Multiplicity**: This is the number of times an eigenvalue is repeated in the characteristic polynomial. For instance, an eigenvalue of multiplicity 2 would appear as \((\lambda - \lambda_i)^2\) in the polynomial. Algebraic multiplicity provides a measure of the roots of the characteristic equation but does not always equal the number of eigenvectors.
- **Geometric Multiplicity**: This refers to the dimension of the eigenspace associated with an eigenvalue. It shows the actual number of linearly independent eigenvectors for an eigenvalue.
It must satisfy the rule that geometric multiplicity is always less than or equal to algebraic multiplicity. Understanding these concepts helps in identifying the complete structure of a matrix and helps discern cases needing generalized eigenvectors.
Characteristic Polynomial
The characteristic polynomial of a matrix is a crucial tool in eigenvalue calculations. It is derived from the determinant of a matrix formed by subtracting \( \lambda \) times the identity matrix from the original matrix.
This can be mathematically expressed as:
  • \( \,\det(A - \lambda I) = 0 \)
This polynomial equips us with all the possible eigenvalues of a matrix, as solving the polynomial equals finding its roots.
The degree of the characteristic polynomial matches the size of the matrix, indicating that an \( n \times n \) matrix could have up to \( n \) eigenvalues. Understanding this polynomial not only provides eigenvalues but also enables the discernment of algebraic multiplicity through repeated roots.
Matrix Diagonalization
Matrix diagonalization is a process to transform a square matrix into a diagonal form while maintaining its core properties. This transformation is possible when a matrix has enough eigenvectors to fill the columns of an invertible matrix \( P \). If \( A \) can be diagonalized, it is expressed as:
  • \( A = PDP^{-1} \)
Here, \( P \) is the matrix with eigenvectors as its columns, and \( D \) is a diagonal matrix with corresponding eigenvalues on its diagonal.
Diagonalization simplifies matrix operations significantly, making powers of matrices easy to compute and solving systems of differential equations more straightforward. However, not every matrix is diagonalizable. Matrices with too few independent eigenvectors, typically when geometric multiplicity is less than algebraic, can't be diagonalized.

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Most popular questions from this chapter

Untersuchen Sie die linearen Gleichungssysteme $$ \left(\begin{array}{llll} 1 & 3 & 1 & 3 \\ 2 & 3 & 1 & 4 \\ 3 & 3 & 0 & 1 \end{array}\right)\left(\begin{array}{c} x \\ y \\ z \\ w \end{array}\right)=\left(\begin{array}{l} 5 \\ 1 \\ 2 \end{array}\right),\left(\begin{array}{llllll} 2 & 3 & 0 & 1 & 0 & 0 \\ 0 & 2 & 5 & 0 & 1 & 0 \\ 3 & 2 & 4 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{c} x \\ y \\ z \\ u \\ v \\ w \end{array}\right)=\left(\begin{array}{r} 8 \\ 10 \\ 15 \end{array}\right) $$ auf Lösbarkeit und ermitteln Sie gegebenenfalls alle Lösungen.

Berechnen Sie die Eigenwerte, Eigenvektoren und gegebenfalls Hauptvektoren der Matrizen $$ A=\left(\begin{array}{llr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 3 & -3 \end{array}\right), B=\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right), C=\left(\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) $$

Untersuchen Sie das lineare Gleichungssystem \(A_{k} \mathbf{x}=\mathbf{b}_{k}\) auf Lösbarkeit in Anhängigkeit der reellen Parameter \(\alpha, \beta\) und ermitteln Sie gegebenenfalls alle Lösungen für $$ A_{1}=\left(\begin{array}{lll} 1 & 3 & 1 \\ 2 & 3 & 1 \\ 3 & 3 & \alpha \end{array}\right), \mathbf{b}_{1}=\left(\begin{array}{l} 5 \\ 5 \\ 5 \end{array}\right) $$A_{2}=\left(\begin{array}{rrr} 1 & 1 & -1 \\ 2 & 3 & \beta \\ 1 & \beta & 3 \end{array}\right), \mathbf{b}_{2}=\left(\begin{array}{l} 1 \\ 3 \\ 2 \end{array}\right)

Untersuchen Sie das lineare Gleichungssystem \(A_{k} \mathbf{x}=\mathbf{b}_{k}\) auf Lösbarkeit und ermitteln Sie gegebenenfalls alle Lösungen für $$ \begin{aligned} &A_{1}=\left(\begin{array}{rrr} 2 & -2 & 0 \\ -1 & 2 & -1 \\ 1 & 0 & -1 \end{array}\right), \mathbf{b}_{1}=\left(\begin{array}{r} 3 \\ -3 \\ 0 \end{array}\right) \\ &A_{2}=\left(\begin{array}{rrrr} 2 & -2 & 1 & 2 \\ -1 & 2 & -1 & -2 \\ 1 & 0 & -1 & 1 \\ 0 & 1 & 3 & 2 \end{array}\right), \mathbf{b}_{2}=\left(\begin{array}{r} 3 \\ -3 \\ 0 \\ 5 \end{array}\right) \end{aligned} $$

Berechnen Sie die Koordinaten des Vektors a \(=(1,2,3)^{T}\) bezügl. der Basis, bestehend aus den Vektoren \(\mathbf{b}_{1}=(1,1,1)^{T}, \mathbf{b}_{2}=(1,2,1), \mathbf{b}_{3}=(0,0,1)^{T}\).

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