Jordan Decomposition: Solve 4x4 Matrix Eqn

[springo]

Homework Statement

I need to find a Jordan decomposition for:
\[ \left( \begin{array}{cccc}<br /> 2 &amp; 0 &amp; 1 &amp; 2 \\<br /> -1 &amp; 3 &amp; 0 &amp; -1 \\<br /> 2 &amp; -2 &amp; 4 &amp; 6 \\<br /> -1 &amp; 1 &amp; -1 &amp; -1 \end{array} \right)\]

Homework Equations

The Attempt at a Solution

I found the eigenvalues: 2 (m=4).
I also found the eigenvectors:
\[ \left( \begin{array}{c}1 &amp; 1 &amp; 0 &amp; 0\end{array} \right)\]\[ \left( \begin{array}{c}-1 &amp; 0 &amp; -2 &amp; 1\end{array} \right)\]
But then I see that (A-2I)² = 0.
So how do I continue?

Thanks a lot.

 

[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= v<sub>2[/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 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; 0 &amp; 0 \\0 &amp; 0 &amp; 2 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 2\end{bmatrix}

If the second is the case, the Jordan Normal Form is
\begin{bmatrix}2 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 2 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 2\end{bmatrix}</sub>