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.

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

Step-by-step explanation:

Given that

N ∈ L(V) is nilpotent

attached below is the required prove

step-by-step explanation:

6 equilateral triangles