I'm stuck to prove Nilpotent Matrix

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A nilpotent matrix A of dimension n*n satisfies the condition A^k=0 for k>=m, indicating that all eigenvalues are zero with multiplicity n and index m or less. The discussion highlights the challenge of proving the converse: if all eigenvalues of A are zero, then A^k must equal zero. The suggested approaches include utilizing Jordan normal form and analyzing the characteristic polynomial to establish this relationship definitively.

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crazygrey
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Hi all,
If a square matrix A of dimension n*n is a nilpotent matrix,i.e, A^k=0 for k>=m iff A has eigenvalues 0 with multiplicity n and index m or less. I did prove by induction if A^k=0 then all the eigenvalues are zero. I'm lost when I want to prove the oppesite, i.e, if all eigenvalues are zero then A^k=0? Please help
 
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Put it in jordan normal form and it all drops out. Alternatively just think about the characteristic poly.
 

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