Reducing a matrix to its Jordan normal form

**arbolis** · September 6th 2010, 01:12 PM

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

**HallsofIvy** · September 6th 2010, 02:11 PM

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

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