How can I accurately find eigenvalues for a Jordan canonical form matrix?

  • Thread starter Thread starter pinguhash
  • Start date Start date
  • Tags Tags
    Eigenvalues
pinguhash
Messages
4
Reaction score
0

Homework Statement


Ok I was working with finding Jordan canonical form...
Here is the matrix I was working on:
| 1 1 1 |
|-1 -1 -1 |
| 1 1 0 |

I am having problem with finding eigenvalues... below is the attempt to solution
I was not getting the right answer. So, when I used online calculator to find the eigenvalue it was comletely different from what I got!

2. The attempt at a solution

|λ-1 , 1 , 1 |
|-1 , λ+1 , -1 |
|1 , 1 , λ |

So, I got values something λ3 = 4

The values from online calculator was λ3 = 0

Please help me in finding how they got eigenvalues all 0.
 
Physics news on Phys.org
You can either do det(A - xI) = 0, or det(xI - A) = 0, where A is your matrix and x I used instead of lambda. You chose to go with the second one. but forgot to change the signs of the non-diagonal entries.

Btw, I think this should go into "Calculus and Beyond". JCF certainly isn't precalculus material :P
 
oh sorry for wrong section, I am new to this forum...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top