# Thread: Reducing a matrix to its Jordan normal form

1. ## Reducing a matrix to its Jordan normal form

I must reduce the following matrix to its Jordan normal form: $\displaystyle M= \begin {bmatrix} 2&1&-1&0&1&0 \\ 0&2&0&1&1&0 \\ 0&0&2&0&1&0 \\ 0&0&0&2&1&0 \\ 0&0&0&0&2&1 \\ 0&0&0&0&0&-1 \end{bmatrix}$.
My attempt: I first calculate its characteristic polynomial which gave me $\displaystyle p(\lambda )= (2-\lambda)^5(-1-\lambda)$.
So there are 2 different eigenvalues, namely $\displaystyle \lamdba _1=2$ and $\displaystyle \lambda _2 =-1$.
Now, I think I have to find the solution of $\displaystyle (M-\lambda _1I)^5 v_1=0$.
I did lots of arithmetics to reach that $\displaystyle (M-\lambda _1I)^5 =\begin {bmatrix} 0&0&0&0&0&-27 \\ 0&0&0&0&0&-27 \\ 0&0&0&0&0&-27 \\ 0&0&0&0&0&-27 \\ 0&0&0&0&0&81 \\ 0&0&0&0&0&-243 \end{bmatrix}$ so that it gives me $\displaystyle v_1=\vec 0$... Mathematica doesn't agree on $\displaystyle (M-\lambda _1I)^5$, it seems I've made some errors in some values in the last column, but the matrix has the same form and therefore $\displaystyle v_1$ is still worth $\displaystyle 0$ which doesn't give me anything useful I believe.
I don't know how to proceed.

2. You don't want to solve $\displaystyle (M- \lambda_1I)^5v=0$. You want to start by solving $\displaystyle (M- \lambda_1I)v= 0$ (such solutions will, of course, also satisfy $\displaystyle (M- \lambda_1I)^5v= 0$). From the Jordan Normal form, it looks like you should be able to find two independent solutions, call them $\displaystyle v_1$ and $\displaystyle v_2$. Those correspond to the two "$\displaystyle \lambda_1$"s on the diagonal with "0" above them. Then look for "generalized eigenvectors" that satisfy $\displaystyle (M- \lamba_1I)v= v_1$ and $\displaystyle (M- \lambda_1I)v= v_2$ (those will also satisfy $\displaystyle (M- \lambda_1I)^5v= 0$).