Multiplicity of Eigenvalues in a 3x3 Matrix

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SUMMARY

The matrix A = \begin {bmatrix} 0 & -1 & 0 \\ 0 & -1 & 0 \\ 0 & 1 & 0 \end{bmatrix} has two real eigenvalues: -1 with multiplicity 1 and 0 with multiplicity 2. The characteristic polynomial derived from cofactor expansion is -\lambda^3 - \lambda^2 = 0, which simplifies to \lambda^2(\lambda + 1) = 0. The reduced row echelon form (RREF) of the matrix is \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, indicating that there are two linearly independent eigenvectors corresponding to the eigenvalue 0.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors
  • Familiarity with matrix operations and RREF
  • Knowledge of characteristic polynomials
  • Proficiency in linear algebra concepts
NEXT STEPS
  • Study the process of finding eigenvalues and eigenvectors in 3x3 matrices
  • Learn about the geometric interpretation of eigenvalues and their multiplicities
  • Explore the implications of eigenvector bases in linear transformations
  • Investigate the relationship between RREF and the dimension of eigenspaces
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Students studying linear algebra, educators teaching matrix theory, and anyone interested in understanding eigenvalue multiplicities in matrices.

icefall5
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Homework Statement


The matrix A = \begin {bmatrix} 0 & -1 & 0 \\ 0 & -1 & 0 \\ 0 & 1 & 0 \end{bmatrix} has two real eigenvalues, one of multiplicity 2 and one of multiplicity 1. Find the eigenvalues and a basis of each eigenspace.

Homework Equations


N/A

The Attempt at a Solution


I've done cofactor expansion to come up with the equation - \lambda^3 - \lambda^2, and that the eigenvalues are therefore -1 and 0, but I don't know how to determine the multiplicity of each. I've looked it up and have gotten nowhere. I can do the rest of the problem, I just don't know how to get these multiplicities.

EDIT: Working on the problem further, and I got the basis for -1. There are supposed to be two vectors for 0, however, but the RREF of the matrix is \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. How does that translate to two eigenvectors?

Thanks in advance for any help!
 
Last edited:
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icefall5 said:

Homework Statement


The matrix A = \begin {bmatrix} 0 & -1 & 0 \\ 0 & -1 & 0 \\ 0 & 1 & 0 \end{bmatrix} has two real eigenvalues, one of multiplicity 2 and one of multiplicity 1. Find the eigenvalues and a basis of each eigenspace.


Homework Equations


N/A


The Attempt at a Solution


I've done cofactor expansion to come up with the equation - \lambda^3 - \lambda^2,
That's not an equation. The equation is - \lambda^3 - \lambda^2 = 0, or ##\lambda^2(\lambda + 1) = 0##
Can you see that one of the eigenvalues has multiplicity 2 and the other has multiplicity 1?
icefall5 said:
and that the eigenvalues are therefore -1 and 0, but I don't know how to determine the multiplicity of each. I've looked it up and have gotten nowhere. I can do the rest of the problem, I just don't know how to get these multiplicities.

EDIT: Working on the problem further, and I got the basis for -1. There are supposed to be two vectors for 0, however, but the RREF of the matrix is \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. How does that translate to two eigenvectors?

Thanks in advance for any help!
 
Ah, yes, I just couldn't simplify that properly; sorry! Thank you!

[STRIKE]But now the second part -- how does that matrix equate to two eigenvectors?[/STRIKE]

I'm an idiot; I figured it out. Thanks again!
 
Last edited:

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