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1 vote
Suppose N in L(V) is nilpotent. Prove that the minimal polynomial fo N is z^m+1, where m is the length of the longest consecutive strong of 1st that appears on teh line directly aoce the diagonal in teh matrix of N with respect to any jordan basis for N.

asked
User Harryngh
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8.0k points

1 Answer

3 votes

Answer: your question is poorly written attached below is the complete and well written question

answer : The minimal polynomial has degree m + 1 hence Z^m+1

Explanation:

Given that

N ∈ L(V) is nilpotent

attached below is the required prove

Suppose N in L(V) is nilpotent. Prove that the minimal polynomial fo N is z^m+1, where-example-1
Suppose N in L(V) is nilpotent. Prove that the minimal polynomial fo N is z^m+1, where-example-2
answered
User Gusaki
by
7.9k points
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