Multiplicity of Eigenvalues in a 3x3 Matrix

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The matrix A has two eigenvalues: -1 with multiplicity 1 and 0 with multiplicity 2, derived from the characteristic polynomial -λ^3 - λ^2 = 0. The user initially struggled to determine the multiplicity of each eigenvalue but later clarified their understanding. They found the basis for the eigenspace corresponding to -1 and were confused about obtaining two eigenvectors for the eigenvalue 0. After further analysis, they resolved their confusion regarding the relationship between the reduced row echelon form and the number of eigenvectors. The discussion highlights the process of finding eigenvalues and their multiplicities in a 3x3 matrix.
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!
 
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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!
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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