Prove that the trace of a nilpotent matrix is always zero and that the trace of any power of that matrix is also zero.

Hello student,
Please find the answer to your question below
LetAbe anilpotent matrix. AssumeAⁿ=0. Letλbe aneigenvalueofA. ThenAx=λxfor some nonzerovectorx. Byinductionλⁿx=Aⁿx=0, soλ=0.
Conversely, suppose that all eigenvalues ofAarezero. Then the chararacteristicpolynomialofA:det(λI−A)=λⁿ. It now follows from the Cayley-Hamilton theorem thatAⁿ=0.
Since thedeterminantis theproductof the eigenvalues it follows that a nilpotent matrix has determinant 0. Similarly, since thetraceof a square matrix is thesumof the eigenvalues, it follows that it has trace 0.
As in nilpotent matrix Aⁿ=0 it also follows that trace(Aⁿ)=0 similarly