Nilpotency of Matrix with one eigenvalue

  • Thread starter geor
  • Start date
35
0
Hello everybody,

I have a question for which I cannot find the answer around,
any help would be really appreciated.

Suppose we have a matrix A of a linear transformation of a vector space,
with only one eigenvalue, say 's'.

My question is: Is the operator (A-sI) nilpotent? ('I' is the identity matrix).

I am trying to understand the proof of the Jordan Canonical Form Theorem and there is
a fuzzy point in my notes about that...

Thanks a lot in advance..

EDIT: I am becoming pretty sure about that, but some confirmation would be great..
 
Last edited:

HallsofIvy

Science Advisor
Homework Helper
41,682
864
I assume you are talking about finite dimensional vector spaces. A linear transformation from a vector space of dimension n to itself can be represented by an n by n matrix. If an n by n matrix, over the complex numbers, has only the single eigenvalue, [itex]\lambda[/itex], then its characteristic equation must be [itex](x- \lambda)^n= 0[/itex]. Since every matrix satisfies its own characteristic equation, [itex](A- \lambda I)^n= 0[/itex].
 
35
0
Oh yes, I see it now! Thanks so much!
 

Related Threads for: Nilpotency of Matrix with one eigenvalue

Replies
2
Views
2K
Replies
7
Views
7K
  • Posted
Replies
2
Views
6K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top