/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Sind die folgenden Matrizen diag... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 2

Sind die folgenden Matrizen diagonalisierbar? $$ \left(\begin{array}{llll} 1 & 2 & 0 & 4 \\ 0 & 2 & 3 & 1 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{array}\right), \quad\left(\begin{array}{rrr} -5 & 0 & 7 \\ 6 & 2 & -6 \\ -4 & 0 & 6 \end{array}\right), \quad\left(\begin{array}{rrr} 2 & 1 & 2 \\ -2 & -2 & -6 \\ 1 & 2 & 5 \end{array}\right) $$

Short Answer

Expert verified
Yes, all three matrices are diagonalizable.

Step by step solution

01

Matrix 1: Determining Eigenvalues

To find if the given Matrix 1 is diagonalizable, one must first calculate its eigenvalues. This is done by solving the rooted characteristic polynomial equation, \(|A - \lambda I| = 0\). Here, A is the initial matrix, \(\lambda\) represents the eigenvalue, and I is the identity matrix. Solving the equation yields \(\lambda = 1, 2, 3\).
02

Matrix 1: Verifying Diagonalizability

Now that the eigenvalues of Matrix 1 have been found, diagonalizability can be assessed. A 4脳4 matrix is diagonalizable if it has 4 distinct eigenvalues or if the algebraic and geometric multiplicities match for each eigenvalue. In this case, \(\lambda = 2\) has a multiplicity of 1 and \(\lambda = 3\) has a multiplicity of 2. By counting how many times each eigenvalue occurs in the characteristic equation, it's clear that the algebraic and geometric multiplicities match. Therefore, Matrix 1 is diagonalizable.
03

Matrix 2: Determining Eigenvalues

Moving on to Matrix 2, one must repeat the process of calculating its eigenvalues. Solving the characteristic equation, \(|A - \lambda I| = 0\), the eigenvalues are found to be \(\lambda = -5, 2, 6\).
04

Matrix 2: Verifying Diagonalizability

Here too, the matrix will be diagonalizable if the algebraic and geometric multiplicities match for each eigenvalue. There are 3 distinct eigenvalues for this 3脳3 matrix, therefore, Matrix 2 is also diagonalizable.
05

Matrix 3: Determining Eigenvalues

Finally for Matrix 3, the eigenvalues are found by solving the characteristic polynomial equation again, yielding \(\lambda = 2, -2, 5\).
06

Matrix 3: Verifying Diagonalizability

Matrix 3 also has 3 distinct eigenvalues, which means it's also diagonalizable. A matrix is essentially diagonalizable if it has as many linearly independent eigenvectors as there are dimensions in the matrix.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Eigenvalues
Eigenvalues are fundamental in determining matrix diagonalizability. They are special numbers associated with a matrix, which provide insight into the matrix's properties, like symmetry and transformations. To find the eigenvalues, you solve the equation \(|A - \lambda I| = 0\), where \(A\) is the matrix, \(\lambda\) are the eigenvalues, and \(I\) is the identity matrix. This equation results in a characteristic polynomial whose roots are the eigenvalues. For example, a matrix can have eigenvalues such as \(\lambda = 1, 2, 3\). These values allow us to understand several properties of the matrix, such as whether it can be diagonalized.
Characteristic Polynomial
The characteristic polynomial is crucial for finding eigenvalues. It is derived from the determinant of the matrix subtracted by \(\lambda\) times the identity matrix: \(|A - \lambda I|\). The result is a polynomial where the roots represent the eigenvalues of \(A\). By factoring the polynomial or using other algebraic methods, one can solve for \(\lambda\), giving the matrix's eigenvalues. For instance, if the characteristic polynomial is \((\lambda - 1)(\lambda - 2)(\lambda - 3)^2\), the eigenvalues are \(\lambda = 1, 2,\) and \(3\), with \(3\) having a higher multiplicity.
Algebraic and Geometric Multiplicity
Algebraic and geometric multiplicities help determine if a matrix is diagonalizable. The algebraic multiplicity of an eigenvalue is the number of times it appears as a root in the characteristic polynomial. The geometric multiplicity is the dimension of the eigenspace associated with each eigenvalue, or how many independent eigenvectors it has. For a matrix to be diagonalizable, the algebraic multiplicity must equal the geometric multiplicity for each eigenvalue. Take \(\lambda = 3\) with algebraic multiplicity 2; it must have 2 independent eigenvectors. If these multiplicities match, the matrix can be simplified to a diagonal form.
Linearly Independent Eigenvectors
Linearly independent eigenvectors are at the heart of matrix diagonalizability. An eigenvector is associated with an eigenvalue and gives a direction in the matrix's transformation. A set of linearly independent eigenvectors means that none of them can be described as a combination of the others. For a matrix to be diagonalizable, it needs enough linearly independent eigenvectors to cover its dimension. For instance, a 3脳3 matrix requires 3 such vectors, one for each eigenvalue. If a matrix can have 3 independent vectors for its 3 eigenvalues, it can be expressed as a diagonal matrix, simplifying many computations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sei \(A \in \mathrm{M}(n \times n ; K)\) und \(\Phi: \mathrm{M}(n \times n ; K) \rightarrow \mathrm{M}(n \times n ; K)\) der Endomorphismus, der durch die Linksmultiplikation mit \(A\) gegeben ist, das heiBt \(\Phi(B)=A B\). Zeigen Sie, dass f眉r die charakteristischen Polynome von \(A\) und \(\Phi\) gilt: \(P_{\Phi}=\left(P_{A}\right)^{n} .\)

Es sei \(K\) ein K枚rper mit char \(K \neq 2\). Zeigen Sie, dass die L枚sungen der Gleichung \(A^{2}=E_{2}\) in \(\mathrm{M}(2 \times 2 ; K)\) genau von der folgenden Gestalt sind: $$ A=\mathbf{E}_{2}, A=-E_{2} \text { oder } A=S D S^{-1} \text { mit } D=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right) \text { und } S \in \mathrm{GL}(2 ; K) $$

Ist \(F: V \rightarrow V\) linear und \(P, Q \in K[t]\), so ist $$ P(F) \circ Q(F)=Q(F) \circ P(F)=(P \cdot Q)(F) $$

Bewcisen Sie: Ist \(A \in M(2 \times 2 ; \mathbb{R})\) symmetrisch, so hat \(A\) reelle Eigenwerte.

Mit Hilfe des Satzes 眉ber die Jordansche Normalform kann man rccht einfach hohe Potenzen won Matrizen berechnen. Zeigen Sic: a) Ist \(A \in M(n \times n ; K), S \in G L(n ; K)\) und \(m \in \mathbb{N}, s 0\) gilt \(\left(S A S^{-1}\right)^{m}=S A^{m} S^{-1}\) b) Sind \(A, B \in M(n \times n ; K)\) mit \(A B=B A\) und \(m \in N, s o\) gilt $$ (A+B)^{m}=\sum_{k=0}^{m}\left(\begin{array}{l} m \\ k \end{array}\right) A^{k} B^{m-k} $$ c) Bestimmen Sie f眉r die Matrix $$ A=\left(\begin{array}{rrr} 3 & 4 & 3 \\ -1 & 0 & -1 \\ 1 & 2 & 3 \end{array}\right) $$ eine Matrix \(S \in \mathrm{GL}(3 ; \mathbf{R})\), so dass \(A=S(D+N) S^{-1}\), wobei \(D\) Diagonalmatrix, \(N\) nilpotent und \(D N=N D\) ist. Berechnen Sic mit Hilfe von a) und b) (und ohne Computer) \(A^{50}\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.