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

# Jordan cannonical form worked example

### Real triple root example with dimension 3

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### sohrab:

2015-12-01 08:06:25DEAR

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### Admin:

2015-12-02 17:44:39Thanks a Lot, Sohrab!

Pleased to be useful.

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