Please check my proof (generalized eigenspaces)

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SUMMARY

The discussion centers on proving that for a linear operator T on a vector space V, if the ranks of the operators (T-λI)m and (T-λI)m+1 are equal for some integer m, then the generalized eigenspace Kλ is equal to the null space N((T-λI)m. The proof builds on earlier results that established the equality of null spaces for consecutive powers of (T-λI). The participant identifies a potential mistake in their reasoning regarding the case when p is greater than or equal to m, questioning whether they can conclude that x is in N((T-λI)m when m equals p.

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Homework Statement


let T be a linear operator on V, and let λ be an eigenvalue of T. prove that if rank((T-λI)m = rank((T-λI)m+1 for some integer m, then Kλ = N((T-λI)m.


Homework Equations


this is part 4 of a 6 part proof, and in the earlier stages, i showed if rank((T-λI)m = rank((T-λI)m+1 for some integer m, then N((T-λI)m = N((T-λI)m+1.


The Attempt at a Solution


let x be in Kλ. by previous result, N((T-λI)m = N((T-λI)k for k≥ m. x is in Kλ, so x is in N((T-λI)p for some integer p. if p≥m, then x is in N((T-λI)m. if p<m, then m = p+t for some t. then (T-λI)p+t(x) = (T-λI)t(T-λI)p(x) = (T-λI)t(0)=0, so x is in N((T-λI)m.
let y be in N((T-λI)m. then y is in N((T-λI)p for some integer p, so y in in Kλ, and the two sets are equal.
 
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well, i see a mistake already, in the "if p≥m" part...
 
can i just say, since (T-λI)p(x)=0, then x is N((T-λI)m if m=p?
 

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