Show that r is repeated root for characteristic equation iff

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To show that r is a multiple root of the minimal polynomial u(x) for the linear operator A, it is necessary to establish that the kernel of (A - rI) is a proper subset of the kernel of (A - rI)². The discussion highlights that the kernel of (A - rI) contains the zero vector and eigenvectors associated with r. It also introduces the concept of generalized eigenvectors, indicating that if v is an eigenvector for r, a vector v2 can be found such that (A - rI)v2 = v. This leads to the conclusion that v2 belongs to the kernel of (A - rI)² but not to the kernel of (A - rI), thus supporting the claim about the multiple root. The relationship between the minimal polynomial and the eigenvalues is emphasized as crucial for the proof.
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Homework Statement


A:B→B a linear operator

Show r is multiple root for minimal polynomial u(x) iff

>$$\{0\}\subset \ker(A - rI) \subset \ker(A - rI)^2$$

note: it is proper subset

Homework Equations



The Attempt at a Solution


Homework Statement



My thought:

I know ker(A−rI) is basically {{0} and {eigenvectors associated with r}}.

what is ker((A−rI)^2) with respect to above and/or r? How is eigenvector of (A−rI)^2 related to that of (L−rI)?
 
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Have you ever heard of a generalized eigenvector?

http://en.wikipedia.org/wiki/Generalized_eigenvector

Suppose v is an eigenvector corresponding corresponding to r. If you could find a vector v2 such that (A - rI) v2 = v. Then:

(A - rI)2 v2 = (A - rI) v = 0​

So v2 \in ker( (A - rI)2). But v2 \notin ker(A - rI).

See if you can use the fact that (x-r)2 divides the minimal polynomial to show that such a v2 exists.
 
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