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Compute the Jordan Canonical form of $\begin{bmatrix}0 & 1 & 0 & 0\\0 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{bmatrix}$


I don't know what to do after computing the characteristic polynomial, which i got $x^4$=0. This would give us an eigenvalue of x=0 of multiplicity 4.

Any thoughts/comments on where to go after this? Thanks so much!

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Huh? Isn't the matrix already a Jordan form? –  user1551 Nov 13 '13 at 9:07

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Next you need to find the minimal polynomial. It is one of $x^2,x^3,x^4$. Use direct verification- here clearly matrix multiplied by itself gives zero, hence the minimal polynomial is $x^2$.Therefore largest Jordan block $J(0,l)$ is of size 2 (that occurs in JNF ) and 4 is the sum of the sizes of these Jordan blocks.

Also the dimension of the eigenspace for an eigenvalue equals the number of Jordan blocks with that eigenvalue. Here there are therefore two Jordan blocks corresponding to the eigenvalue zero. So basically the matrix is already in Jordan Normal Form.

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