# Jordan canonical form for a matrix

How do I find the Jordan canonical form and its transitions matrix of this matrix?

\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}

The characteristic polynomial is (x+1)*(x-1)^3 and the eigenvectors are for x=1 we have (0,0,0,1)', (0,1,1,0)', (1,0,0,0)' and for the x=-1 we have (0,-1,1,0)'.

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HINT: For each eigenvalue, the geometric multiplicity agrees with the algebraic multiplicity. – vadim123 Dec 8 '13 at 21:00
Check the minimal pol. of the matrix is $\;(x-1)(x+1)\;$ and thus it is diagonalizable, what makes its JCF pretty boring...and simple. – DonAntonio Dec 8 '13 at 21:05

$$J = \left[ \begin{array}{rrrr} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right]$$