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
for this, we make
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
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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Thanks 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?
Absolutly, both anwers are correct. Only you must take care with the change of basis matrix.
I agree with Necip, but this is a great example!
Thanks 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.