Nilpotency of Matrix with one eigenvalue

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SUMMARY

The discussion centers on the nilpotency of a matrix A with a single eigenvalue 's' in the context of linear transformations. It is established that if A is an n by n matrix over the complex numbers with only one eigenvalue, then the operator (A - sI) is indeed nilpotent. This conclusion is supported by the characteristic equation (x - s)^n = 0, which leads to (A - sI)^n = 0, confirming nilpotency. The conversation highlights the relevance of this concept in understanding the Jordan Canonical Form Theorem.

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Mathematicians, students of linear algebra, and anyone studying matrix theory or the Jordan Canonical Form will benefit from this discussion.

geor
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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..
 
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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, \lambda, then its characteristic equation must be (x- \lambda)^n= 0. Since every matrix satisfies its own characteristic equation, (A- \lambda I)^n= 0.
 
Oh yes, I see it now! Thanks so much!
 

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