More on Jordan canonical forms

Dear Professor Kim,

I was just doing some practice by redoing some of the 2201 homework. On sheet 5, Q5, part (g) asks us to find the Jordan Canonical Form. ch_T(X)=(X-6)^8 and null (T-6 id)= 7

Am i correct to believe the null=7 tells us that there should be 7 Jordan blocks? in that case should the answer be diag(J_2(6), 6,6,6,6,6,6)? The solutions I printed from your site stated that the answer is diag(J_7(6),6). Am I correct to think that there is an error there? Could you please also confirm the rest of the answers are correct there for Q5?

Also, in the lecture 14/11 you went over an example to calculate the Jordan Canonical Form given the ch_A, m_A of a matrix with two eigenvalues. I am still finding it quite difficult to use the method you presented in your lecture and there are no past paper solutions to look at for another example. Would you kindly present us with another example by either choosing a question from a past paper or from any
other source you have.

Your help is much appreciated.

Thank you

Reply:

You are absolutely right about 5 (g). Your answer is correct and there was an error in the sheet. Thanks very much for pointing this out. All the others seem to be correct.

To see an example with two eigenvalues involving no additional work, let’s just `combine’ two parts of that exercise. Suppose

ch_T(X)=(x-3)^3(x-2)^3

and

m_T(X)=(x-3)^2(x-2)

Then we know that we have a 6 by 6 matrix and the JCF will have two 3 by 3 blocks, one for each eigenvalue, since 3 is the multiplicity of the two eigenvalues in the characteristic polynomial. To find the shape of each block, argue exactly in the same way as when the two parts were separate. For the block corresponding to eigenvalue 3, the multiplicity two in the minimal polynomial says that the largest Jordan block has size two. So the corresponding 3 by 3 block will be

diag(J_2(3), 3)

Corresponding to the eigenvalue 2, the largest Jordan block has size 1 because that is the multiplicity of 2 in the minimal polynomial. So the 3 by 3 block there is

diag(2,2,2)

Finally, the JCF for the original matrix then becomes

diag(J_2(3),3,2,2,2)

Sometimes people wonder about the ordering of the two blocks. In this respect, the JCF is not unique. You could also have written

diag(2,2,2,J_2(3),3)

Some people like the convention of listing the blocks according to the size of the eigenvalues (if they are real).

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: