The converse is not true. A simple counterexample is when A=[1−11−1],
A2=[1−11−1][1−11−1]=[0000]
The method of proof is identical, so only the case for upper triangular matrices is discussed.
Proof
We prove this by mathematical induction.
When n=1, it holds.
Let A be a 1×1 upper triangular matrix.
A=[0]
It is trivially nilpotent.
Assume it holds for n=k, then for n=k+1, it holds.
Let A be a (k+1)×(k+1) upper triangular matrix. Then, for the k×k upper triangular matrix B, A is represented as the following block matrix.
A=B[00⋯0]a1k+1a2k+1⋮akk+1[0]=[BO1kC[0]]
Then, calculating the power of A yields the following.
A2A3⋮Ap+1=[BO1kC[0]][BO1kC[0]]=[B2O1kBC[0]]=[B2O1kBC[0]][BO1kC[0]]=[B3O1kB2C[0]]=[Bp+1O1kBpC[0]]
Let Bp=Okk. Then since Ap+1=Ok+1k+1, the n=k+1 upper triangular matrix is nilpotent.
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Stephen H. Friedberg, Linear Algebra (4th Edition, 2002), p512 ↩︎