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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
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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

1 Answer 1

Since you have the geometric multiplicity equalling the algebraic multiplicity for each eigenvalue, then the JCF form of this matrix will only be entries on the diagonal. This is because the number of blocks to each eigenvalue is equal to its geometric multiplicity. Hence,

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

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