## Yet more JCF

Professor Kim,
Thanks for the answer of the previous question. Let me present you with anotherquestion. I am more concerned of questions where the nullity and rank are given, where the m_A and ch_A do not give enough information to calculate the Jordan Canonical form.

Suppose we have the following (from 2006 exam paper):

ch_A(X)= (X-1)^10(X-2)^10, m_A(X)=(X-1)^6(X-2)^4,
null(A-I)=4, rank(A-2I)=16 and null((A-2I)^2)=8

How should we approach the question here?

Thank you

Firstly, ranks can be converted to nullities by a standard formula. Now, what information is to be had from the nullities of $(A-\lambda I)^i$? This is really where you have to focus on the discussion of the text. Look at the array of Jordan basis elements on, say, page 2 of the `12-11-07 Notes’ from the course webpage. The shape of this array completely determines the contribution of a given eigenvalue $\lambda$ to the JCF. How?

(1) Each column corresponds to a Jordan block for $\lambda$;

(2) The height of each column is exactly the size of the corresponding block.

Now, back to the nullities. The main facts are:

(a) The nullity of $(A-\lambda I)$ is the number of linearly independent eigenvectors for the eigenvalue $\lambda$, which is exactly the number of elements in the bottom row;

(b) The nullity of $(A-\lambda I)^2$ is exactly the number of elements in the bottom two rows ;

(c) The nullity of $(A-\lambda I)^3$ is exactly the number of elements in the bottom three rows ;

and so on.

Examine the problem again with that in mind. Also, read the relevant section from the notes again and verify for yourself that what I wrote above is exactly what’s written there.