Question

Is it true that if ab is nilpotent then ba is nilpotent matrix?

Answer 1

9down voteaccepted

Hint: I use this theorem: If ∀i≥1∀i≥1 trac Ci=0Ci=0, then CC is nilpotent.

You can easily prove by induction that trac (Ci)=0Ci)=0 for all i≥1i≥1.

Edit1: Theorem :∀i≥1∀i≥1 trac Ci=0Ci=0 iff C is nilpotent.

Proof: CC is a real matrix but you assume AA is a complex matrix and f(x)=(x−a1)(x−a2)...(x−an)f(x)=(x−a1)(x−a2)...(x−an) is its characteristic polynomial in the complex field. You can prove that trac(CkCk)=∑ni=1aki∑i=1naik by induction, and if ∀k∈N∀k∈N trac(CkCk)=∑ni=1aki=0∑i=1naik=0 thenai=0ai=0. Hence, f(x)=xnf(x)=xn, so An=0An=0 and it is shown that C is nilpotent.