Trace of powers of a nilpotent matrix

The answer of Qiaochu Yuan: Here is an argument that does not involve Newton’s identities, although it is still closely related to symmetric functions.[r]

FOR WORDPRESS - TRACE OF POWERS OF A NILPOTENT MATRIX

HO `ANG PHI D ˜UNG †

Abstract Trace of powers of a nilpotent matrix

1 Introduction This note, I copied from Yoshi on MathStackExchange:

LetA be an n×n complex nilpotent matrix Then we know that because all eigenvalues of A must be 0, it follows that tr(An) = for all positive integers n.

What I would like to show is the converse, that is, if tr(An) = for all positive integers

n, then A is nilpotent I tried to show that must be an eigenvalue of A, then try to show that all other eigenvalues must be equal to However, I am stuck at the point where I need to show that det(A) =

May I know of the approach to show thatA is nilpotent?

The answer of Alvarez:

If the eigenvalues ofAare λ1, .,λn, then the eigenvalues ofAkareλk1, .,λkn It follows

that if all powers ofA have zero trace, then λk

1 +· · ·+λkn = 0,∀k≥1

Using [Newton’s identities] to express the elementary symmetric functions of the λi’s in

terms of their power sums, we see that all the coefficients of the characteristic polynomial of

A (except that of greatest degree, of course) are zero This means thatA is nilpotent

The answer of JBC:

Assume that for allk = 1, , n, tr(Ak) = where A is an×n matrix We consider the

eigenvalues in C

Suppose A is not nilpotent, so A has some non-zero eigenvalues λ1, , λr ¡br /¿ Let ni

the multiplicity ofλi then

   

  

n1λ1+· · ·+nrλr =

n1λr1+· · ·+nrλrr =

So we have      

λ1 λ2 · · · λr

λ21 λ22 · · · λ2r

λr

1 λr2 · · · λrr

            n1 n2 nr       =       0       But det      

λ1 λ2 · · · λr λ21 λ22 · · · λ2r

λr

1 λr2 · · · λrr

     

=λ1· · ·λrdet

       

1 · · ·

λ1 λ2 · · · λr λ21 λ22 · · · λ2r

λr1−1 λr2−1 · · · λr−1

r         = (Vandermonde)

So the system has a unique solution which is n1 = .=nr = Contradiction

The answer of Qiaochu Yuan: Here is an argument that does not involve Newton’s identities, although it is still closely related to symmetric functions Write

f(z) = X

k≥0

zktr(Ak) =

n

X

i=1

1 1−zλi

where λi are the eigenvalues of A As a meromorphic function, f(z) has poles at the

reciprocals of all of the nonzero eigenvalues of A Hence if f(z) = n identically, then there are no such nonzero eigenvalues

The argument using Newton’s identities, however, proves the stronger statement that we only need to require tr(Ak) = for 1≤k ≤ n Newton’s identities are in fact equivalent to

the identity

f(z) = n− zp 0(z) p(z) wherep(z) =Qn

i=1(1−zλi) To prove this identity it suffices to observe that

logp(z) =

n

X

i=1

log(1−zλi)

and differentiating both sides gives

p0(z)

p(z) =

n

X

i=1 −λi

1−zλi

(The argument using Newton’s identities is also valid over any field of characteristic zero.) Department of Scientific Fundamentals, Posts and Telecommunications Institute of Tech-nology, Office A2, Fl.10, Km10 Nguyen Trai Rd., Ha Dong District, Hanoi, Vietnam

E-mail address: dunghp@ptit.edu.vn