Does there exist a complex matrix $A$ with shape $n\times n$ such that $\operatorname{Tr}(A^k)=k$ for all natural $k$?

Question

$\quad$ Does there exist a complex matrix $A$ with shape $n\times n$ such that $\operatorname{Tr}(A^k)=k$ for all natural $k$?

$\quad$ I solve this problem using charasteristic polynomial of $A$ and prove that the all eigenvalues of $A$ is $1$ then $\operatorname{Tr}(A^k)=n$ so that there is not exist matrix with such properties.

$\quad$ I am haunted by the thought that there is a shorter and more elegant solution. If you find or have a more elegant and short solution, please show me. $\quad\chi_A(t)=\det(tI-A)=t^n-c_{n-1}t^{n-1}-\ldots-c_1t-c_0\Rightarrow \chi_A(A)=0\Rightarrow$ $$ A^n=c_{n-1}A^{n-1}+\ldots+c_1A+c_0I\Rightarrow $$ $$ A^{n+k}=c_{n-1}A^{n+k-1}+\ldots+c_1A^{k+1}+c_0A^k\Rightarrow $$ $$ \operatorname{Tr}(A^{n+k})=c_{n-1}\operatorname{Tr}(A^{n+k-1})+\ldots+c_1\operatorname{Tr}(A^{k+1})+c_0\operatorname{Tr}(A^k)\Rightarrow $$ $$ n+k=c_{n-1}(n+k-1)+\ldots+c_1(k+1)+c_0k\Rightarrow $$ $$ c_{n-1}+\ldots+c_0=1\Rightarrow \chi_A(1)=0\Rightarrow $$ $$ A\sim\begin{bmatrix} 1 & u\\ 0 & B \end{bmatrix}\Rightarrow $$ $$ \operatorname{Tr}(B^k)=k-1 $$ $$ \ldots $$ All eigenvalues of $A$ is equal to $1$

Answer 1

$$\DeclareMathOperator{\Tr}{Tr}$$ Work over any $\mathbb{Q}$-algebra, not just $\mathbb{C}$. The assumption can be organized into the formal power series $$\Tr\frac{A}{1-tA}=\frac{1}{(1-t)^2}\text{.}$$ Formally integrating and exponentiating using Newton's identities or the Jacobi formula gives $$\det(1-tA)^{-1} = \exp\left(\frac{t}{1-t}\right)\text{,}$$ i.e., $$\det(1+tA) = \exp\left(\frac{t}{1+t}\right)\text{.}$$ The left side terminates in degree $n$, but the right doesn't—a contradiction. (In a different generality, if $A$ is a trace-class operator on a separable Hilbert space, $\det(1+tA)$ is holomorphic in $t$, so we still have a contradiction.)

Answer 2

Note that any non-real eigenvalues must come in conjugate pairs (all traces are $\in \mathbb R$ so Newton's Identities imply a characteristic polynomial $\in\mathbb R[x]$). Now consider the Hankel matrix $H_A$ whose component $i,j$ equals $\text{trace}\big(A^{i+j-1}\big)$ $\big[\text{ref}$ $\forall k\in\mathbb{Z},\sum_{i=1}^n\lambda_i^k=(\sum_{i=1}^n\lambda_i)^k$ implies at most one number of $\{\lambda_i\}$ doesn't equal $0$? $\big]$

$\text{rank}(H_A)=2$ since its $j$th column is equal to the first column + $(j-1)\cdot\mathbf 1$ $\implies A$ has exactly $2$ distinct non-zero eigenvalues per the link.

The leading $2\times 2$ principal submatrix has determinant $1\cdot 3 - 2\cdot 2=-1$ so $H_A=V^T DV\not \succeq \mathbf 0\implies A$ has a non-real eigenvalue (consider the leading $2\times 2$ principal minor of $H_{A^2}$ to rule out $A$ having a negative and positive eigenvalue) $\implies$ $A$ has $\lambda$ and $\overline \lambda$ each with multiplicity $m$. Thus the original post reads $2m\cdot\text{Re}\big(\lambda^k\big)= m\cdot \lambda^k + m\cdot(\overline \lambda)^k =\text{trace}\big(A^k)=k$ for all $k\in \mathbb N\implies \lambda \in \mathbb R$ since $\lambda^k$ is in the right half plane for all $k$; thus $\lambda$ is both real and non-real and that is a contradiction.

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sketch of a different approach:
For any $\mathbb F$ where $\text{char }\mathbb F = 0$ (or for positive characteristic we impose the usual restriction of $\text{char }\mathbb F \gt n$)
if such an $A$ exists then there must be some minimal $n$ (obviously $n\geq 2$) where $A$ exists. So we focus on this minimal case; this implies $A$ is invertible. Now compute the same thing two different ways:

(i.) $\text{rank}\big(H_A\big)=2$ as above so $A$ has exactly 2 unique non-zero eigenvalues
[technically we work over a splitting field whenever eigenvalues are mentioned]
(ii.) consider the Hankel matrix whose component $i,j$ equals $\text{trace}\big(A^{i+j-2}\big)$, i.e. the one here: If $A$ has distinct eigenvalues, then all matrices close enough to $A$ also have distinct eigenvalues . If $n\geq 3$ then we can check that the leading $3\times 3$ principal submatrix has determinant $-n\neq 0$ hence that matrix has rank (at least) $3$ hence $A$ has (at least) $3$ unique eigenvalues, all of which are non-zero which contradicts (i.). Conclude: $n=2$.

$ \begin{pmatrix} 2 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3& 4 \\ \end{pmatrix} =\begin{pmatrix} \text{trace}\big(I_2\big) & \text{trace}\big(A^1\big) & \text{trace}\big(A^2\big) \\ \text{trace}\big(A^1\big) & \text{trace}\big(A^2\big) & \text{trace}\big(A^3\big) \\ \text{trace}\big(A^2\big) & \text{trace}\big(A^3\big)& \text{trace}\big(A^4\big) \\ \end{pmatrix}= V^T V$
where $V$ is $2\times 3$ but the matrix on the left has determinant $=-2$
$\implies 3=\text{rank}\begin{pmatrix} 2 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3& 3 \\ \end{pmatrix} =\text{rank}\big(V^T V\big)\leq \text{rank}\big( V^T\big)\leq 2$
which is a contradiction

alternative finish after (ii.)

Checking $A$'s first two traces implies the Companion Matrix $\left[\begin{matrix}0 & \frac{1}{2}\\1 & 1\end{matrix}\right]$ generates the OP's traces, but
$\text{trace}\left(\left[\begin{matrix}0 & \frac{1}{2}\\1 & 1\end{matrix}\right]^3\right) =\frac{5}{2}\neq 3$

Answer 3

Complex Analysis Proof:
Let $A$ have characteristic polynomial $p(z)$ and spectral radius $R\geq 1$ since $\text{trace}\big(A^k\big)$ is bounded away from $0$. Then in the annulus $\big\{z: R \lt \vert z\vert\lt \infty\big\}$ the rational function $r(z) = \frac{p'(z)}{p(z)}=\sum_{k=1}^n\frac{1}{z-\lambda_k}$, has Laurent Series
$ = \sum_{k=1}^\infty \text{trace}\big(A^{k-1}\big)\cdot z^{-k}= n z^{-1} + \sum_{k=2}^\infty (k-1)\cdot z^{-k}$
(check Cauchy's Integral Formula and the fact that $r\big(\infty\big)=0$)

This implies we have analytic $f: \big(0, R^{-1}\big)\longrightarrow \mathbb C$ given by
$f(z) = r\big(z^{-1}\big) = z\cdot\big(n + \sum_{k=2}^\infty (k-1)\cdot z^{k-1}\big)= n\cdot z + z^2\cdot \frac{d}{dz}\Big(\frac{1}{1-z}\Big)$
where $0$ is a removable singularity that we've defined over. $f$ has a pole at some $\vert z\vert =R^{-1}$ but it inherits a radius of convergence of $1$ from $\frac{1}{1-z}\implies R^{-1}\geq 1\implies R^{-1}=1=R$.

$\text{Triangle inequality }\implies n\geq \Big\vert\text{trace}\big(A^{n+1}\big)\Big\vert=n+1$ and that is a contradiction

Takeaway:
The underlying problem for the OP is that $\big\vert\text{trace}\big(A^k\big)\big\vert\to \infty$ but it does so too slowly allowing $f$ to be analytic in $B\big(0,1\big)$. In particular $\big\vert\text{trace}\big(A^k\big)\big\vert\to \infty\implies \sum_{k=1}^\infty \text{trace}\big(A^{k}\big)\cdot z^{k}$ cannot converge absolutely in the unit disc

Answer 4

Let $B = (A - A^2)$ then $B^n$ has trace $0$ for all $n \geq 2$.

To see this let

$(x - x^2)^n = a_nx^n + \dots + a_{2n}x^{2n}$

then trace of $B^n$ $=$ $na_n + \ldots + 2na_{2n}$

Note that RHS is just derivative of $(x - x^2)^n$ evaluated at $x = 1$, which is a root of multiplicity $>$ $1$ whenever $n > 1$ hence the derivative $0$.

In particular if $C = B^2$ then $C$ and all its powers have trace $0$ $\implies$ $C$ is nilpotent $\implies$ $B$ is nilpotent $\implies$ $A$ and $A^2$ have equal trace which is a contradiction.

Answer 5

The trace is the sum of the EV. So if $A$ has EV $\lambda_i$, where we count each EV multiple times according to its algebraic multiplicity, then $A^k$ has as its EV $\lambda_i^k$. Now $f(x):=Tr(A^x)=\sum_{i=1}^n \lambda_i^x$ is a sum of exponentials, and as such only linear if it sums to 0.