Jordan cannonical form worked example

Real triple root example with dimension 3

Let's make a worked example of Jordan form calculation for a 3x3 matrix.

Let's the matrix



Calculate the roots of characteristic polynomial, ie calculate the eigenspace AX=λX, this is given for the equation system A-λI=0



Therefore, we have the λ=3 triple multiplicity eigenvalue.

We need to know the dimension of the eigenspace generated by this eigenvalue, ie, calculate dim [Ker (A-3I)], to do that, we solve the system

(A-3I)X=0

So, it appears



Thus we obtain a single eigenvector



Thus the dimension of the eigenspace is 1. Therefore, A is not diagonalizable, and we know that the Jordan form of A is



We calculate the basis of eigenvectors, for now we only have one, the v1, that is, we have to calculate two more eigenvectors. For this, we apply the theory, ie, we calculate a vector, v2 such that.

(A-3I)v2 = v1, ie solve the system



Now we have a second eigenvector, we now calculate the third one, v3, such that

(A-3I)v3 = v2, ie solve the system



We're done because we have to

A = PJP-1

Where



And P is those matrix formed the eigenvectors placed as columns, ie






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Post from other users

sohrab:

2016-12-01 08:06:25
DEAR

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Admin:

2016-12-02 17:44:39
Thanks a Lot, Sohrab!
Pleased to be useful.
Enjoy

Regards,
Admin

abdul rehman:

2016-12-17 14:23:05
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