Jordan basis and jordan normal form

[specialnlovin]

Let
A=

2 -1 2 1 
1 0 3 1
-2 1 0 1
-1 0 0 3

the characteristic polynomial of A is (x-1)³(x-2)
find the minimal polynomial, jordan basis, and jordan normal form
I know the minimum polynomial is (x-1)(x-2), but I am not sure how to find the nordan basis and jordan normal form

 

[micromass]

The eigenvalues are 1 and 2.
The algebraic multiplicities of the eigenvalues tells how many times each eigenvalues there are on the diagonal. Here, the number 1 will appear 3 times on our diagonal and the number 2 will appear 1 time on our diagonal.

Now, the most important thing is the size and the numbers of the Jordan blocks.
The size of the largest Jordan block of an eigenvalue, is the exponent of this eigenvalue in the minimal polynomial. So, in our example here, what are the sizes of the largest Jordan blocks of the eigenvalues 1 and 2??

 

[specialnlovin]

so the largest sizes of the jordan blocks for 1 and 2 are 1x1 blocks

 

[micromass]

Indeed. But this means that the Jordan normal form is a diagonal matrix! And in particular, the matrix is diagonizable!

Thus the Jordan normal form can easily be written down.
The Jordan basis is, in this case, just the basis for diagonalization (i.e. a basis of eigenvectors)

 

[specialnlovin]

how do we find the basis of eigenvectors though? when i did (A-I) i got only one vector

 

[micromass]

First determine the eigenspaces of 1 and 2.

 

[specialnlovin]

so is the eigenspace of 1 the span of (2, 3, 0, 1)^t? and the eigenspace of 2 the span of (1, 1, 1, 1)^t? but then how do i get the eigenvectors from that?

 

[micromass]

Hmm, I recalculated. It seems your minimal polynomial is wrong. It is not true that

$$(A-I)(A-2I)=0$$

What is the correct minimal polynomial??

 

[specialnlovin]

(a-i)³(a-2i)

 

[specialnlovin]

For V₁(1) i found the vector v₁=(2, 3, 0, 1)^T
for V₂(1) i found the vector v₂=(-1, -1, 1 0)^T
but I cannot figure out v₃

 

[micromass]

Ok, so the correct minimal polynomail is $$(x-1)^3(x-2)$$.
So, what are the sizes of the largest Jordan blocks of 1 and 2?

 

[specialnlovin]

the largest jordan block of 1 is 3x3 and 2 is 1x1.
so the eigenspace of 1 is the span ((2, 3, 0, 1)^T, (-1, -1, 1, 0)^T, (0, 1, 0, 0)^T)
and the eigenspace of 2 is the span of (1, 1, 1, 1)^T right?

 

[micromass]

No these eigenspaces are incorrect. The vectors you mention aren't even eigenvectors...

The next step you need to do is to determine bases for

$$Ker(A-I), Ker[(A-I)^2], Ker[(A-I)^3], Ker(A-2I)$$

This will help us determine the Jordan basis.

If you're confused about what I'm doing. Check the following site www.ms.uky.edu/~lee/amspekulin/jordan_canonical_form.pdf This gives some examples of calculations of the Jordan normal form...