How do I find the Jordan canonical form of this 4x4 matrix?

Question

A = $\begin{bmatrix} 1 & -1 & 0 & -1\\ 0 & 2 & 0 & 1\\ -2 & 1 & -1 & 1\\ 2 & -1 & 2 & 0\\ \end{bmatrix}$

I've been given this matrix - I'm supposed to find its Jordan form as well as matrix M such that: $ A = MJM^{-1}$.

So far, what I have is the eigenvalues: 1, 1, 1, -1. I have three eigenvectors (which I realize is not enough):

A-1I ~ Gauss Jordan ~ $\begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$ $ \qquad \Rightarrow v_1 = \begin{pmatrix} -1 \\ 0 \\ 1\\ 0\\ \end{pmatrix}$ $ \quad v_2 = \begin{pmatrix} 0 \\ -1 \\ 0\\ 1\\ \end{pmatrix}$

A+1I ~~ $\begin{bmatrix} 1 & 1 & 0 & 0\\ 0 & 3 & 0 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$ $ \qquad \Rightarrow v_3 = \begin{pmatrix} 1 \\ -1 \\ -3\\ 3\\ \end{pmatrix}$

This is all I've been able to find. I need to either find a fourth vector in order to find matrix M, or a way to find matrix J and then compute M.

Answer 1

If you had four eigenvectors, you'd be able to diagonalize the matrix. Evidently this matrix is not diagonalizable, but of course it still has a Jordan form. You need to find the "generalized eigenspaces", i.e. the kernels of not just the $(A-\lambda I)$, but also $(A-\lambda I)^n$ for all $n \geq 1$. Since you're only missing one generalized eigenvector here, you'll only need to look at $n=2$ to find your $v_4$.

Answer 2

There is also the matter of consistency. The extra vector, I am going to call it $w,$ has $(A-I)^2 w = 0,$ however $(A-I)w \neq 0.$ Then we take $v = (A-I)w,$ which is automatically an eigenvector. The third and fourth columns of $P$ are $v,w.$ I took column vector $w = (0,0,0,1)^T,$ where the $T$ means transpose. Then $v = (A-I)w = (-1,1,1,-1)^T.$ Next, you take an eigenvector, call it $u,$ that is independent of $v.$ I thought $u = (1,1,-1,-1)^T$ would be nice. Finally, the left hand column is an eigenvector for $-1,$ call it $t= (1,-1,-3,3)^T$

We make the matrix $P = (t,u,v,w) ,$ find $P^{-1},$ and check that $J = P^{-1}AP $ is really in Jordan form