 # Jordan cannonical form worked example

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

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 # Was useful? want add anything?

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