# Jordan form hard problem

I'm trying to find a matrix $P$ such that $J=P^{-1}AP$, where $J$ is the Jordan Form of the matrix:

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

$p(x)=x(x-1)^{4}$ is the characteristic polynomial. Then, the eigenvalues are $(1, 1, 1, 1, 0 ,0 ,0 ,0)$

I'm trying to find the eigenvectors by this way $(A-I)z=w$ for the eigenvectors with eigenvalue $1$ and $Ax=y$ for the eigenvectors with eigenvalue $0$. But this is impossible.

Can somebody help me please?

Thanks for your help.

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How did you calculate the characteristic polynomial? Haven't you made row/column transformations on the matrix? Based on those, $P$ might be recognizable. – Berci May 5 '13 at 23:50
@Berci I calculate the characteristic polynimial by $det(A-xI)$ its easy because $det=tr(A-xI)$ – user63192 May 5 '13 at 23:57

I think there is an issue with your characteristic polynomial. Maybe you want to enclose both terms in the $4^{th}$ power.

The CP is:

$$\lambda^8-4 \lambda^7+6 \lambda^6-4 \lambda^5+\lambda^4 = (\lambda-1)^4 \lambda^4$$

This leads to two eigenvalues $\lambda_1 = 1, \lambda_2 = 0$ with algebraic multiplicity 4.

For each, we arrive at two eigenvectors and two generalized eigenvectors (first two found normal way, last two generalized for each eigenvalue), that are:

• $v_1 = (0, 0, 0, 0, 0, -1, 1, 0), ~ \lambda_1 = 1$
• $v_2 = (1, 0, 0, 0, 0, 0, 0, 0), ~ \lambda_2 = 1$
• $v_3 = (0, 0, 0, 0, -1, 1, 0, 0), ~\lambda_3 = 1$
• $v_4 = (0, 0, 0, 0, 0, 1, 0, 0), ~\lambda_4 = 1$
• $v_5 = (0, 0, 0, -1, 1, 0, 0, 0), ~ \lambda_5 = 0$
• $v_6 = (0, -1, 1, 0, 0, 0, 0, 0), ~ \lambda_6 = 0$
• $v_7 = (0, -1, 0, 1, 0, 0, 0, 0), ~\lambda_7 = 0$
• $v_8 = (0, 1, 0, 0, 0, 0, 0, -1),~ \lambda_8 = 0$

From the eigenvalues and eigenvectors, we can write the Jordan Normal Form as:

$A = P \cdot J \cdot P^{-1} = \begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & -1 & -1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 1 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}0 & -1 & -1 & -1 & 0 & 0 & 0 & -1\\ 0 & -1 & -1 & 0 & 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & -1 & -1 & -1 & -1 & 0 & 0 & -1\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}$

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Nice work + 1 $\quad \ddot\smile\quad$ – amWhy May 6 '13 at 0:11
Yes, indeed, lots of typing that would require...! – amWhy May 6 '13 at 0:15
Thanks for your help :D – user63192 May 6 '13 at 0:51
@user63192: No problem, make sure to work out the eigenvectors and understand how to derive them - especially the generalized ones. Regards – Amzoti May 6 '13 at 0:56
@BabakS.: Thank you my friend, I even know there is a way to better format the negative signs, but was too tired to look it up! Have a great day! – Amzoti May 6 '13 at 14:29

Here is aJordan form computed by maple

$$\left[ \begin {array}{cccccccc} 1&1&0&0&0&0&0&0\\0 &1&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0 \\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&1&1&0 &0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0 &0&1\\ 0&0&0&0&0&0&0&0\end {array} \right] .$$

The eigenvalues are given by

$$[0,0,0,0,1,1,1,1]$$

and the corresponding eigenvectors are the columns of the matrix

$$\left[ \begin {array}{cccccccc} 0&0&0&0&1&0&0&0\\ 0 &-1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0 \\ -1&0&0&0&0&0&0&0\\1&0&0&0&0&0&0 &0\\ 0&0&0&0&0&-1&0&0\\ 0&0&0&0&0& 1&0&0\\ 0&0&0&0&0&0&0&0\end {array} \right]$$

Now, you should be able to construct $P$.

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But the matrix J, must have four ones in the diagonal , no? – user63192 May 6 '13 at 0:13
Thanks you very much – user63192 May 6 '13 at 0:51
@user63192: You are very welcome. – Mhenni Benghorbal May 6 '13 at 0:53