A square matrix B is called nilpotent if Bm 0 for some m N

A square matrix B is called nilpotent if Bm = 0 for some m N. Prove that if B is a nilpotent n×nmatrix,thenBn =0.
Hint : If Bm = 0 for some m N, it would help to consider the least such natural number m, so that Bm = 0 but Bm1 = 0. (Does such m exist for every nilpotent matrix?)

Solution

A matrix B is called nilpotent if there exists some positive integer, n, such that xⁿ = 0.

If B^m = 0 for some m N, it would help to consider the least such natural number m, so that B^m = 0. Thus we have to prove that det(B) must be 0.

If B is nilpotent, then there exists some integer m >= 1 such that B^m = 0, so then det(B^m) = 0
The determinant of the product is equal to the product of the determinants, namely det(A*B) = det(A)*det(B).
So, det(B^m) = (det(B))^m = 0.

Therefore, det(B) must be 0. Thus, proved.