If the trace of the matrices $A^k$ are equal to the size of $A$ why the spectrum of $A$ is $\{1\}$ Let $A$ a matrix in $\mathcal{M}_n(\mathbb C)$ such that $\operatorname{Tr}(A^k)=n$ for all $k\in\{1,\ldots,n\}$. I have to prove that $\operatorname{Sp}(A)=\{1\}$. I denoted $\lambda_1,\ldots,\lambda_n$ the eigenvalues of $A$ with multiplicity so the hypothesis leads to the system of equations: $\lambda_1^k+\cdots+\lambda_n^k=n$ for all $k$. I can see that $\lambda_1=\cdots=\lambda_n=1$ is a trivial solution of the system but it's not obvious for me why it would be the unique solution. Any suggestion?
 A: You can do it by induction. In fact, let $P(X) = \prod_{i=1}^n \left(X-\lambda_i\right) = \sum_{k=0}^na_kX^k$. You will have :
$$0 = \sum_{i=1}^n P\left(\lambda_i\right) = \sum_{k=0}^n a_k \sum_{i=1}^n \lambda_i^k = n \times \sum_{k=0}^n a_k = n \times P(1)$$ So $P(1) = 0$ there is $i$ such that $\lambda_i = 1$. You can assume that is $\lambda_n$ and you will have :
$$\sum_{i=1}^{n-1} \lambda_i^k = n-1^k = n -1,\quad\forall k\in\left\{0,\ldots,n-1\right\}$$ By induction $\lambda_{n-1} = \ldots =\lambda_1 = 1$.
A: In order to establish that the characteristic polynomial of $A$ is $(1-\lambda)^n$, we have to show (binomial expansion) that the generic coefficient of this characteristic polynomial is
$$c_k=(-1)^k \binom{n}{k}. \tag{1}$$
Do you know the Faddeev-Leverrier algorithm for obtaining the coefficients of the characteristic polynomial of a matrix $A$ ?
It is based on the recursive formula given in the Wikipedia reference:
$$\displaystyle c_{n-m}=-{\frac {1}{m}}(c_{n}\mathrm {tr} A^{m}+c_{n-1}\mathrm {tr} A^{m-1}+...+c_{n-m+1}\mathrm {tr} A)=-{\frac {1}{m}}\sum _{k=1}^{m}c_{n-m+k}\mathrm {tr} A^{k} \tag{2}$$
Can you now use (2) to establish (1) by induction using the following formula for binomial coefficients:
$$\displaystyle \sum _{j=0}^{k}(-1)^{j}{\binom {n}{j}}=(-1)^{k}{\binom {n-1}{k}} ? \tag{3}$$
A: A big idea is to convert your problem into a more common one: that of nilpotence.
$B := I-A$
(motivation: you are told that $\lambda_i=1$ for all $i$ and need to prove it. If this is true, then $I-A$ has all eigenvalues of zero -- proof: use similarity transform to upper triangularize $A$. The rest of the post just flushes this out in terms of Binomial Theorem.)
$B^k=\big(I-A\big)^k = \sum_{j=0}^k (-1)^j\binom{k}{j}A^j$
(where we defined $B^0:= I$ for notational ease)
$\implies $ for $k\in\big\{1,2,...,n\big\}$
$\text{trace}\big(B^k\big) $
$= \text{trace}\big(\sum_{j=0}^k (-1)^j\binom{k}{j}A^j\big)$
$= \sum_{j=0}^k (-1)^j\binom{k}{j}\text{trace}\big(A^j\big)$
$= n\cdot \sum_{j=0}^k (-1)^j\binom{k}{j}1$
$= n\cdot (1-1)^k$
$=0$
Thus $\text{trace}\big(B^k\big)=0$ for $k\in\big\{1,2,..,n\big\}$ which implies $B$ is nilpotent (ref e.g. Traces of all positive powers of a matrix are zero implies it is nilpotent -- the Vandermonde Matrix approach is easiest) i.e. all eigenvalues are zero which implies $A=I-B$ has all eigenvalues of one.
A: We can use two properties :

*

*Trace is sum of eigenvalues.

*Eigenvalues transform as the matrix does.

1 and 2 together give us: $$\sum_{i=1}^n {\lambda_i}^k = n, \forall k$$
Which complex numbers $\lambda_i$ could this hold for?
Let us consider vector $\lambda = [\lambda_1,\cdots,\lambda_n]$ with standard inner product $<.,.>$. Just the straight forward "dot" product.
This means for example (k=1,k=2) : $n=<\lambda,1> = <\lambda,\lambda>$
This can only be true for $\lambda$ being the $1$ vector.
A: For $1\le k \le n,$ let $p_k=x_1^k+\cdots+x_n^k.$ Since the ring
$\mathbb Q[p_1,\cdots, p_n]$ equals all rational symmetric functions,
it contains every elementary symmetric function. Each $p_k$ has the same value whether evaluated at $(1, \cdots, 1)$ or at $(\lambda_1, \cdots, \lambda_n).$ Thus, the eame holds for each elementary symmetric function. Therefore,
$$(X-1)^n=(X-\lambda_1)\cdots (X-\lambda_n)$$
