Jordan Normal Form of Matrices

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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