Questions about Jordan Normal Form

[Sudharaka]

Hi everyone, :)

Here's a question I found on a Wiki book.

Question: The matrix \(T\) is \(5\times 5\) with the single eigenvalue 3. The nullities of the powers are: \((T-3I)\) has nullity two, \((T-3I)^2\) has nullity three, \((T-3I)^3\) has nullity four, and \((T-3I)^4\) has nullity five. Find the Jordan Normal form from the given date.

So I can understand that since the nullity of \((T-3I)\) is two there are two Jordan blocks. However I still don't get how exactly we can calculate the sizes of those Jordan blocks from the above data. Can anybody guide me through this process please? :)

 

[I like Serena]

Hi everyone, :)

Here's a question I found on a Wiki book.

Question: The matrix \(T\) is \(5\times 5\) with the single eigenvalue 3. The nullities of the powers are: \((T-3I)\) has nullity two, \((T-3I)^2\) has nullity three, \((T-3I)^3\) has nullity four, and \((T-3I)^4\) has nullity five. Find the Jordan Normal form from the given date.

So I can understand that since the nullity of \((T-3I)\) is two there are two Jordan blocks. However I still don't get how exactly we can calculate the sizes of those Jordan blocks from the above data. Can anybody guide me through this process please? :)

Suppose you have a 3x3 matrix with the single eigenvalue 3 and 1 Jordan block.
Then the nullity of $(T-3I)$ is 1. That is, there is 1 eigenvector.
When you take a look at $(T-3I)^2v = 0$, there are 2 solutions: the original eigenvector is still a solution, and we get another generalized eigenvector from the root space.
In other words the nullity of $(T-3I)^2$ is 2.
Finally $(T-3I)^3 = 0$ gives access to the entire root space. It has nullity 3.Back to your 5x5 matrix with 2 eigenvectors for the eigenvalue 3.
What are the possible configurations of the Jordan blocks?
Suppose there are 2 Jordan blocks: one of size 2 and another one of size 3.
How many generalized eigenvectors are added if you look at $(T-3I)^2$?

 

[Sudharaka]

Suppose you have a 3x3 matrix with the single eigenvalue 3 and 1 Jordan block.
Then the nullity of $(T-3I)$ is 1. That is, there is 1 eigenvector.
When you take a look at $(T-3I)^2v = 0$, there are 2 solutions: the original eigenvector is still a solution, and we get another generalized eigenvector from the root space.
In other words the nullity of $(T-3I)^2$ is 2.
Finally $(T-3I)^3 = 0$ gives access to the entire root space. It has nullity 3.Back to your 5x5 matrix with 2 eigenvectors for the eigenvalue 3.
What are the possible configurations of the Jordan blocks?
Suppose there are 2 Jordan blocks: one of size 2 and another one of size 3.
How many generalized eigenvectors are added if you look at $(T-3I)^2$?

Sorry for not replying earlier. Since the nullity of \((T-3I)\) is two we should have two Jordan blocks. There are two possible configurations;

1) One Jordan block of size 1 and the other of size 4.

2) One Jordan block with size 2 and the other of size 3.

Am I correct up to this point? :)

 

[I like Serena]

Sorry for not replying earlier. Since the nullity of \((T-3I)\) is two we should have two Jordan blocks. There are two possible configurations;

1) One Jordan block of size 1 and the other of size 4.

2) One Jordan block with size 2 and the other of size 3.

Am I correct up to this point?

Yep!

 

[Sudharaka]

Yep!

Okay so the next step is to size of the Jordan blocks. So we have to use the formula,

\[N_{k}(\lambda)=r_{k+1}(\lambda)-2r_{k}(\lambda)+r_{k-1}(\lambda)\]

where \(N_{k}(\lambda)\) is the number of Jordan blocks of size \(k\). \(r_{k}(\lambda)\) is the rank of the matrix \((T-\lambda I)^k\).

Am I correct? Or is there a easier method? :)

 

[I like Serena]

Okay so the next step is to size of the Jordan blocks. So we have to use the formula,

\[N_{k}(\lambda)=r_{k+1}(\lambda)-2r_{k}(\lambda)+r_{k-1}(\lambda)\]

where \(N_{k}(\lambda)\) is the number of Jordan blocks of size \(k\). \(r_{k}(\lambda)\) is the rank of the matrix \((T-\lambda I)^k\).

Am I correct? Or is there a easier method? :)

Sounds difficult and I do not recognize it.
Alternatively, you could take another look at my post #2...

 

[Sudharaka]

Sounds difficult and I do not recognize it.
Alternatively, you could take another look at my post #2...

This was a method that I picked up when going through my lecture notes. :) Anyway I shall try to proceed the way you suggest in post #2.

Suppose you have a 3x3 matrix with the single eigenvalue 3 and 1 Jordan block.
Then the nullity of $(T-3I)$ is 1. That is, there is 1 eigenvector.
When you take a look at $(T-3I)^2v = 0$, there are 2 solutions: the original eigenvector is still a solution, and we get another generalized eigenvector from the root space.
In other words the nullity of $(T-3I)^2$ is 2.
Finally $(T-3I)^3 = 0$ gives access to the entire root space. It has nullity 3.

Back to your 5x5 matrix with 2 eigenvectors for the eigenvalue 3.
What are the possible configurations of the Jordan blocks?
Suppose there are 2 Jordan blocks: one of size 2 and another one of size 3.
How many generalized eigenvectors are added if you look at $(T-3I)^2$?

I have given the possible Jordan block configurations in my post #3. Now when I look at \((T-3I)^2\) I see that one generalized vector is added. Hope I am correct. :)

Now what's the next step? :)

 

[Sudharaka]

Hi everyone, :)

Here's a question I found on a Wiki book.

Question: The matrix \(T\) is \(5\times 5\) with the single eigenvalue 3. The nullities of the powers are: \((T-3I)\) has nullity two, \((T-3I)^2\) has nullity three, \((T-3I)^3\) has nullity four, and \((T-3I)^4\) has nullity five. Find the Jordan Normal form from the given data.

So I can understand that since the nullity of \((T-3I)\) is two there are two Jordan blocks. However I still don't get how exactly we can calculate the sizes of those Jordan blocks from the above data. Can anybody guide me through this process please? :)

I think I understood this now. I am going to write down the complete answer again. It would be really nice if somebody could confirm what I did was correct.

Let \(c_{k}\) be the number of Jordan blocks of size greater than or equal to \(k\). And \(c_{k}\) is given by,

\[c_{k}=\mbox{dim Ker}(T-3I)^k-\mbox{dim Ker}(T-3I)^{k-1}\]

So we know that,

Number of Jordan blocks of size greater than or equal to 2 = 1

Number of Jordan blocks of size greater than or equal to 3 = 1

Number of Jordan blocks of size greater than or equal to 4 = 1

Since \(\mbox{dim Ker}(T-3I)=2\) we know that there are two possible configurations for the Jordan normal form. The two Jordan blocks could be size 1 and 4 or they could be of sizes 2 and 3.

Hence the Jordan blocks should be,

\[J_{1}(3)=(3)\mbox{ and }J_{4}(3)=\begin{pmatrix}3&1&0&0\\0&3&1&0\\0&0&3&1\\0&0&0&3\end{pmatrix}\]

Therefore the Jordan normal form of the matrix \(T\) should be,

\[\begin{pmatrix}3&0&0&0&0\\0&3&1&0&0\\0&0&3&1&0\\0&0&0&3&1\\0&0&0&0&3\end{pmatrix}\]

 

[I like Serena]

Looks good!

 

[Sudharaka]

Looks good!

Thanks very much for all the help you provided. Now I understand the computational method of Jordan Normal form perfectly. What is left to do is to perhaps understand the theoretical parts which are there in my lecture notes, especially the formula given above. :)