Let's make another worked example of Jordan form calculation for a 3x3 matrix, now with
a only eigenvalue with triple and eigenspace spanned with 2 dimension.

Let's the matrix

The fist we calculate the roots of characteristic polynomial:

therefore we have λ=3 triple (algebraic) multiplicity eigenvalue.

Calculate the dimension of eigenspace corresponding to this eigenvalue, ie, calculate

dim[Ker(A-3I)]

for this, we make

(A-3I)X=0

and have the linear equations system

So, we have two eigenvectors

so, 2 is the dimension of eigenespace, therefore A is not diagonalizable and
Jordan cannonical form is

Calculate the eigenvector basis, using the method 1 seen in the theory
(see theory in section "Cannonical Jordan form")
,ie, calculate v3 such us

(A-3I)v3 = k1 v1 + k2 v2

them, we have the system

we have now the eigenvectors basis

we are done because

A = PJP-1

where

and P is the change of basis matrix, formed by eigenvectors placed as column vectors, ie

# Jordan cannonical form worked example

### Real triple root example with dimension 3 and eigenspace spanned by the eigenvalue with dimension 2.

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### Post from other users

### Necip:

2019-11-26 19:20:30Thanks for example. By the way, my calculator shows Jordan matrix as a ((3 1 0)(0 3 0)(0 0 3) with a little difference from your solution. Are both of these answer correct?

### Carlos:

2020-01-17 18:10Hi Necip.

Absolutly, both anwers are correct. Only you must take care with the change of basis matrix.

Regards

Carlos

### Josh:

2020-12-01 04:00:37I agree with Necip, but this is a great example!

### Carlos:

2020-12-02 02:15:37Thanks a lot Josh.

Really he is correct, both answers are true, you only have to change the changing basis matrix formed by the eigenvectors, it must be reordered too.

Regards

Carlos

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