# Jordan Normal Form of Matrices

1. Feb 7, 2010

### victoria13

1. The problem statement, all variables and given/known data

Find the Jordan Normal Form of

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

2. Relevant equations

3. 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

2. Feb 7, 2010

anyone?

3. Feb 7, 2010

### Dick

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.

4. Feb 7, 2010

### victoria13

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

using my v2= (2 0 -1)

i get a matrix that doesnt multiply to get a jordan form....

5. Feb 7, 2010

### Dick

(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.