# Thread: Algebra, Problems For Fun (19)

1. ## Algebra, Problems For Fun (19)

A standard notation: By $\displaystyle M_n(R)$ we mean the set of all $\displaystyle n \times n$ matrices with entries from some set $\displaystyle R.$

Problem: Let $\displaystyle A \in M_n(\mathbb{C}).$ Suppose that $\displaystyle \text{tr}(A)=\text{tr}(A^2)=\cdots = \text{tr}(A^n)=0.$ Prove that $\displaystyle A^n=\bold{0}.$

Remark: The above result remains true if we replace $\displaystyle \mathbb{C}$ with any field $\displaystyle F$ of characteristic 0. It is not necessarily true if $\displaystyle \text{char} F \neq 0.$ A counter-example is: $\displaystyle A=\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \in M_2(\mathbb{F}_2).$

2. Using the fact that the characteristic polynomial of an nxn matrix is a monic polynomial with coefficients that can be expressed in terms of it's trace and the traces of it's powers upto the power n we get that

$\displaystyle \chi_A(X) = \pm X^n$ as all the traces are 0.

Then the result follows from the Cayley Hamilton Theorem.

I couldn't think of a way of proving it that didnt rely on these two results but I'm sure there is a nice way.

thanks, pomp.

3. Originally Posted by pomp
Using the fact that the characteristic polynomial of an nxn matrix is a monic polynomial with coefficients that can be expressed in terms of it's trace and the traces of it's powers upto the power n we get that

$\displaystyle \chi_A(X) = \pm X^n$ as all the traces are 0.

Then the result follows from the Cayley Hamilton Theorem.

I couldn't think of a way of proving it that didnt rely on these two results but I'm sure there is a nice way.

thanks, pomp.
let $\displaystyle \lambda_1, \cdots , \lambda_n$ be the eigenvalues of $\displaystyle A.$ we know that every matrix over $\displaystyle \mathbb{C}$ is similar to some triangular matrix. so $\displaystyle A=PBP^{-1},$ for some triangular matrix $\displaystyle B.$

then $\displaystyle 0=\text{tr}(A^k)=\text{tr}(PB^kP^{-1})=\text{tr}(B^k)=\lambda_1^k + \cdots + \lambda_n^k,$ for all $\displaystyle 1 \leq k \leq n,$ which gives us $\displaystyle \lambda_1=\cdots=\lambda_n=0.$ therefore $\displaystyle A^n=\bold{0},$ by Cayley-Hamilton.