Show that r is repeated root for characteristic equation iff

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SUMMARY

The discussion focuses on proving that a scalar \( r \) is a repeated root of the minimal polynomial \( u(x) \) of a linear operator \( A: B \to B \) if and only if the kernel conditions \( \{0\} \subset \ker(A - rI) \subset \ker(A - rI)^2 \) hold true. The relationship between the kernels of \( (A - rI) \) and \( (A - rI)^2 \) is explored, particularly in the context of generalized eigenvectors. The existence of a vector \( v_2 \) such that \( (A - rI)v_2 = v \) for an eigenvector \( v \) is established, demonstrating that \( v_2 \) belongs to \( \ker((A - rI)^2) \) but not to \( \ker(A - rI) \).

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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 [itex]\in[/itex] ker( (A - rI)2). But v2 [itex]\notin[/itex] 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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