# Homework Help: Jordan normal form problem

1. Jan 2, 2010

### springo

1. The problem statement, all variables and given/known data
Find the Jordan normal form of:
$$\begin{bmatrix} -2 & 3 & 1\\ 0 & -4 & -1\\ 1 & 2 & -3 \end{bmatrix}$$

2. Relevant equations

3. The attempt at a solution
I found the eigenvalues: -3 (x3)
I found a single eigenvector: [ 2 , -1 , 1 ]
But now I thought of finding the nullspaces of (M + 3I)2 and (M + 3I)3 (where is the matrix we're studying) but they're empty.

2. Jan 3, 2010

### Dick

You've already shown that the nullspace of (M+3I) is not empty since you have an eigenvalue of -3. Then how can the nullspaces of (M+3I)^2 and (M+3I)^3 be empty?? That would be silly. In fact, (M+3I)^3=0 since it satisfies it's characteristic polynomial. I suggest you try that calculation again.

Last edited: Jan 3, 2010
3. Jan 3, 2010

### HallsofIvy

Since -3 is a triple root, the characteristic equation must be (M+ 3)^3= 0. And, since every matrix satisfies its own characteristic equation, (M+ 3I)^3v= 0 for every vector v.
There must exist vectors, u, such that Mu= 3u and you have found one. Since the eigenvectors do not span the space, there must exist vectors, v, such that Mv is NOT 0 but $(M+ 3I)^2 u= 0$. Since (M+3I)0= 0 certainly, look for a vector such that (M+ 3I)v= [2, -1, 1]. Then you will have $(M+ 3I)^2v= (M+ 3I)[2, -1, 1]= 0$. Once you have found that v, look for w such that $(M+ 3I)w= v$.

4. Jan 3, 2010

### springo

Oh, this is bad. I was using Mathematica for my calculations, it turns out using 2 doesn't do a real matrix multiplication, it does a dot product or something similar. My bad...

Now I took as eigenvectors:
- (M + 3I)2 : [ 1 , 0 , 1 ]
- (M + 3I)3 : [ 0 , 0 , 1 ]
So that:
$$P=\begin{bmatrix} 2 & 1 & 0\\ -1 & 0 & 0\\ 1 & 1 & 1 \end{bmatrix}$$

But doing P·J·P-1 ≠ M
However taking [ 1 , 0 , 0 ] instead of [ 0 , 0 , 1 ] works, why is that?

5. Jan 6, 2010

### springo

Bump for help! :)

6. Jan 6, 2010

### Dick

I'm really not sure what your are doing. What's J? (M+3I)=0. Anything is an eigenvector of that.

7. Jan 8, 2010

### springo

J is the Jordan normal form of M and P is the matrix so that P·J·P-1 = M.

8. Jan 8, 2010

### HallsofIvy

it's not clear to me how you got [0, 0, 1]. As I said before, you want a "v" such that $$(M+3i)^2v= [2, -1, 1]$$. That means you have to solve
$$\begin{bmatrix}1 & 3 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$$$= \begin{bmatrix}2 \\ -1 \\ 1\end{bmatrix}$$.
That gives the equations x+ 3y+ z= 2 -y- z= -1, and x+ 2y= 1. The second equation is the same as z= -y+ 1 and the third is the same a x= -2y+ 1. Taking y= 0 would give v= (1, 0, 1) as you have.

Now look for u such that (M+ 3I)u= v which means you must have
$$\begin{bmatrix}1 & 3 & 1 \\ 0 & -1 & -1 \\ 1 & 2 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$$.
That gives the equations x+ 3y+ z= 1, -y-z= 0. and x+ 2y= 1. The second equation is the same as z= -y and the third equation x= -2y+1. Then the first equation is satisfied for all y. If you take y= 0, you get [1, 0, 1] the same as before, but if you take y= 1, you get u= [-1, 1, -1].

Let P be the matrix having those vectors as columns,
$$P= \begin{bmatrix}2 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & 1 & -1\end{bmatrix}$$
and you will find that
$$P^{-1}AP= \begin{bmatrix}-3 & 1 & 0 \\0 & -3 & 1 \\ 0 & 0 & -3\end{bmatrix}$$.

9. Jan 8, 2010

### springo

The thing is I was taking for the first column of P, an eigenvector of M. For the second one, a vector that is the nullspace of (M+3I)2 but is not an eigenvector of M. For the third one, a vector that is in the nullspace of (M+3I)3 (which is R3) but not in the eigenspace of (M+3I)2. Isn't that a right thing to do?
I mean: (M+3I)3 = 0 so (M+3I)3·[ 0 , 0 , 1 ] = 0.

Last edited: Jan 8, 2010