Linear Algebra / Geomtric Multiplicity

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SUMMARY

The discussion centers on the eigenvalues and multiplicities of the 2x2 matrix A, defined as A = [[3/2, 1/2], [-1/2, 1/2]]. The eigenvalue identified is 1, with an algebraic multiplicity of 2 and a geometric multiplicity of 1. The geometric multiplicity is confirmed through the kernel calculation of (A - I), which indicates one degree of freedom. Additionally, the discussion includes the need to find matrices P and J for the Jordan canonical form, with preliminary eigenvector calculations provided.

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  • Understanding of eigenvalues and eigenvectors
  • Knowledge of algebraic and geometric multiplicity
  • Familiarity with Jordan canonical form
  • Ability to perform matrix operations and kernel calculations
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  • Study the process of finding eigenvectors for repeated eigenvalues
  • Learn about Jordan canonical form and its applications
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Students and professionals in linear algebra, particularly those studying eigenvalues, eigenvectors, and matrix transformations. This discussion is beneficial for anyone preparing for exams or working on matrix theory applications.

Wildcat
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Homework Statement


Let A = a 2x2 matrix row 1 [3/2 ½ ] row 2 [-½ ½]
a) Find the eigenvalue(s) of A
b) Find the algebraic multiplicity of each eigenvalue.
c)Find the geometric multiplicity of each eigenvalue.



Homework Equations





The Attempt at a Solution


I found the eigenvalue to be 1 with an algebraic multiplicity of 2.
I'm not sure about the geometric multiplicity. I think it is 1.
When I look at the dim(ker(A-1I)) I get ½x1 + ½x2=0 which I think means you have 1 degree of freedom therefore the geometric multiplicity is 1.
Can someone tell me if I'm doing this right?
 
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That's right. 1 is a double eigenvalue but it only corresponds to a single eigenvector.
 
Dick said:
That's right. 1 is a double eigenvalue but it only corresponds to a single eigenvector.

Thanks! I also have to find P and J such that A=PJP^-1 where J is the Jordan canonical form of A, but I'm not finished with that.
 
Wildcat said:
Thanks! I also have to find P and J such that A=PJP^-1 where J is the Jordan canonical form of A, but I'm not finished with that.

I know I need to find the eigenvector corresponding to the eigenvalue 1, I found it to be
[1 -1] but I need another one. Do I choose one where (A-1I)v‡0, can I choose [1 0] for the other??
 
Wildcat said:
I know I need to find the eigenvector corresponding to the eigenvalue 1, I found it to be
[1 -1] but I need another one. Do I choose one where (A-1I)v‡0, can I choose [1 0] for the other??

if I can do that, I get P=[1 1] [-1 0] J=[1 ½] [0 1] P^-1=[0 -1] [1 1] (these are 2x2 matrices)


which does = A [3/2 ½] [-½ ½] ?
 

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