# Jordan Decomposition

1. Jan 10, 2009

### springo

1. The problem statement, all variables and given/known data
I need to find a Jordan decomposition for:
$$$\left( \begin{array}{cccc} 2 & 0 & 1 & 2 \\ -1 & 3 & 0 & -1 \\ 2 & -2 & 4 & 6 \\ -1 & 1 & -1 & -1 \end{array} \right)$$$

2. Relevant equations

3. The attempt at a solution
I found the eigenvalues: 2 (m=4).
I also found the eigenvectors:
$$$\left( \begin{array}{c}1 & 1 & 0 & 0\end{array} \right)$$\left( \begin{array}{c}-1 & 0 & -2 & 1\end{array} \right)$$$
But then I see that (A-2I)2 = 0.
So how do I continue?

Thanks a lot.

2. Jan 11, 2009

### HallsofIvy

If 2 is an eigenvalue of multiplicity 4, then (A- 2I)4= 0. There must exist v such that (A- 2I)v= 0 which is the same as Av= 2v: v is an eigenvector corresponding to eigenvalue 2. You say you have found two independent eigenvectors, v1 and v2. But since the multiplicity is 4, there must now exist a vector v such that (A- 2I)v is NOT 0 but (A- 2I)3v= 0. But (A- 2I)3v= (A-2I)3(A- 2I)v= 0. Since (A- 2I)v1= 0 and (A- 2I)v2= 0m that is the same (A- 2I)3v= v1 or (A- 2I)3v= v2[/sup]. If only one of those has a solution, say v3, then there must be a solution to (A- 2I)2v= v3[/sup].
If the first is the case, the Jordan Normal Form is
$$\begin{bmatrix}2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2\end{bmatrix}$$

If the second is the case, the Jordan Normal Form is
$$\begin{bmatrix}2 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2\end{bmatrix}$$

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