Jordan Normal Form of Matrices

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Homework Help Overview

The discussion revolves around finding the Jordan Normal Form of a given matrix. The matrix in question has specific eigenvalues and eigenvectors that are being analyzed in the context of Jordan form transformation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the eigenvalues and eigenvectors obtained, with one participant attempting to compute (A+I)^2 and encountering a zero matrix. Others suggest the need for three vectors that satisfy specific conditions related to the Jordan form.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem of finding the Jordan basis. There is an exploration of different vectors and their relationships, but no consensus has been reached on the final form or method.

Contextual Notes

Participants note the challenge of selecting appropriate vectors outside the eigenspace and the implications of having (A+I)^2 equal to zero, which affects the dimensionality of the eigenspace.

victoria13
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Homework Statement



Find the Jordan Normal Form of

3,0,8
3,-1,6
-2,0,-5


Homework Equations





The Attempt at a Solution



I got the eigen values as lambda=-1

found 2 eigen vectors:

-2
0
1

and

0
1
0

now when i try and do (A+I)^2 I get a zero matrix so cannot seem to find the last column of the P matrix to then calculate P-1AP=J
 
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anyone?
 
If (A+I)^2=0 then ANY vector satisfies (A+I)^2. What you really want to get to the Jordan form is three vectors satisfying (A+I)v1=0, (A+I)v2=0, (A+I)v3=v2. Start by picking v3 outside of the eigenspace spanned by {v1,v2}. That determines the eigenvector v2. Now pick v1 to be any other linearly independent eigenvector.
 
Dick said:
If (A+I)^2=0 then ANY vector satisfies (A+I)^2. What you really want to get to the Jordan form is three vectors satisfying (A+I)v1=0, (A+I)v2=0, (A+I)v3=v2. Start by picking v3 outside of the eigenspace spanned by {v1,v2}. That determines the eigenvector v2. Now pick v1 to be any other linearly independent eigenvector.

ok I am not quite sure what you mean. when I set (A+I)v2=v3

using my v2= (2 0 -1)

i get a matrix that doesn't multiply to get a jordan form...
 
(A+I)v2=0 if v2=(2,0,-1). You have a two dimensional eigenspace spanned by {(-2,0,1),(0,1,0)}. You want to pick three vectors for the basis that satisfy the relations I gave in the last post. Since you want (A+I)v3=v2, you want to pick a v3 OUTSIDE of that eigenspace. Now define v2=(A+I)v3. Since (A+I)^2=0 that will automatically be in the eigenspace. Now pick the final v1 to be another linearly independent vector in the eigenspace. That's a Jordan basis. Just start doing it. You'll see.
 

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