Question

prove that an nxn nilpotent matrix always has det=0

Answer 1

I know that at some point A^n=0, and therefore the determinant is zero, but I must also have to prove that the matrices before it becomes zero have a determinant of zero

Answer 2

and det(0)=0=A^n....and A^n is connected to A somehow

Answer 3

use the fact det[A^n] = [det A]^n

Answer 4

ok, so that would conclude the proof, because only 0^n can give zero

Answer 5

just like x^n = 0, the only solution is x must be 0

Answer 6

k thanks!