Jordan Forms and Eigenproblems

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In summary: I'm not sure how to solve (a). I could try to do so, but I'm not sure if that's the right way to go about it. So can you solve (a) now?
  • #1
shaon0
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Homework Statement


See attachment.

The Attempt at a Solution


In a) ii); The use of a chain diagram is required but I have no idea how to produce one. As for i); I have no idea how to do this.

In b); (B+5I)v=(1,2,-1) and (B+5I)^2v=0. The eigenvalues are 5 (with multiplicity 2) and 2 (w/ AM of 1). Eigenvectors are (-1,-2,-1) and (-2,-2,1). For iii); I row reduced (B+5I) to get basis of {(0,1,2);(3,-1,1)} but I'm not sure that this is the way to go. I'm completely lost with iv) as I can't get P or J. But i'd think P contains the eigenvectors on the columns and J can be deduced from here by solving. I think v) I can do.
 

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  • #2
Hi shaon0!

Not really sure what a chain diagram is.
Sounds like a nifty way to construct the generalized eigenvectors.

As for (a), how much do you know about constructing Jordan normal forms, and about constructing generalized eigenvectors?

For (b) you need to consider that (B+5I)^2v=0, which makes v a generalized eigenvector, and it makes v1=(B+5I)v a regular eigenvector.

How did you find the eigenvalue 2?
Actually, it should be -2.
Did you deduce it based on the trace of matrix B?

The matrix J is the Jordan normal form related to your 2 eigenvalues, for which you get 2 blocks.
Do you know how to define the blocks?
The matrix P is the matrix of the 3 generalized eigenvectors.
 
  • #3
I like Serena said:
Hi shaon0!

Not really sure what a chain diagram is.
Sounds like a nifty way to construct the generalized eigenvectors.

As for (a), how much do you know about constructing Jordan normal forms, and about constructing generalized eigenvectors?

For (b) you need to consider that (B+5I)^2v=0, which makes v a generalized eigenvector, and it makes v1=(B+5I)v a regular eigenvector.

How did you find the eigenvalue 2?
Actually, it should be -2.
Did you deduce it based on the trace of matrix B?

The matrix J is the Jordan normal form related to your 2 eigenvalues, for which you get 2 blocks.
Do you know how to define the blocks?
The matrix P is the matrix of the 3 generalized eigenvectors.

Yes, sorry -2 is the other eigenvalue. It was a typo and yes i used the trace. J2(5)oJ1(-2) is the jordan form. Which I realize now. I'm not sure how to find the generalised eigenvectors. Are the three eigenvectors I have found, the generalised ones? What makes an eigenvector generalised?
 
  • #4
shaon0 said:
Yes, sorry -2 is the other eigenvalue. It was a typo and yes i used the trace. J2(5)oJ1(-2) is the jordan form. Which I realize now. I'm not sure how to find the generalised eigenvectors. Are the three eigenvectors I have found, the generalised ones? What makes an eigenvector generalised?

Good!
And yes, the three vectors you found are the generalized eigenvectors.
Note that one of them is not an eigenvector.

The generalized eigenvectors are also the vectors that are not eigenvectors themselves, but that are a sort of 2nd or higher order eigenvectors.
That is, you find them from (B-λI)r=0, where r is the order of 2 or higher.

The matrix P is the matrix with the generalized eigenvectors as columns.

Take a look at this wiki page:
http://en.wikipedia.org/wiki/Jordan_normal_form
It explains in more detail what generalized eigenvectors are and how you find them.
 
  • #5
I like Serena said:
Good!
And yes, the three vectors you found are the generalized eigenvectors.
Note that one of them is not an eigenvector.

The generalized eigenvectors are also the vectors that are not eigenvectors themselves, but that are a sort of 2nd or higher order eigenvectors.
That is, you find them from (B-λI)r=0, where r is the order of 2 or higher.

The matrix P is the matrix with the generalized eigenvectors as columns.

Take a look at this wiki page:
http://en.wikipedia.org/wiki/Jordan_normal_form
It explains in more detail what generalized eigenvectors are and how you find them.

Oh ok, right I've got it. Your explanations are very clear. Hopefully now I'll pass my linear algebra exam. Thanks again.
 
  • #6
shaon0 said:
Oh ok, right I've got it. Your explanations are very clear. Hopefully now I'll pass my linear algebra exam. Thanks again.

So can you solve (a) now?

(Since understanding about generalized eigenvectors is a prerequisite for it. :wink:)
 

FAQ: Jordan Forms and Eigenproblems

What is a Jordan form?

A Jordan form is a specific type of matrix representation that can be used to solve eigenproblems. It takes the form of a diagonal matrix with blocks along the diagonal, where each block represents a distinct eigenvalue and its corresponding eigenvectors.

What is the significance of Jordan forms in linear algebra?

Jordan forms are important in linear algebra because they allow for the simplification of eigenproblems, making them easier to solve. They also provide insight into the structure and behavior of linear transformations.

How are Jordan forms related to eigenvalues and eigenvectors?

Jordan forms are closely related to eigenvalues and eigenvectors, as they are used to represent and solve eigenproblems. The eigenvalues of a matrix are the values along the diagonal of the Jordan form, and the eigenvectors are the corresponding blocks in the matrix.

Can any matrix be transformed into a Jordan form?

Not all matrices can be transformed into a Jordan form. A matrix must meet certain conditions, such as being square and having distinct eigenvalues, in order for it to have a Jordan form. In some cases, a matrix may only have a partial Jordan form.

How are Jordan forms used in applications?

Jordan forms have a variety of applications in fields such as physics, engineering, and computer science. They are used to solve eigenproblems and analyze linear transformations, and can also be applied in areas such as data compression and signal processing.

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