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