Higher powers of a matrix's relation with its trace Let $A=[a_{ij}]$, where $a_{ij}=u_{i}v_{j}, 1 \leq i \leq n$ and $1\leq j \leq n$ and $u_i,v_j$ belong to $R$ satisfies $A^5=16A$. Find trace(A).
I denoted U as a column matrix having values u1,u2,...,un. And V a row matrix having values v1,v2,...,vn. So that A=UV.
But I am not able to proceed further. Evaluating A^5 would be very tedious so I think that I am missing the trick in this question.
Also backtracking from the answer, I feel $A^5=(trace(A))^4A$. Is there any easy way to prove this?
 A: Your matrix $A$ can be written more accurately as $A=uv^T$ where $u,v$ are both thought of as column vectors, so that $u$ is $n\times 1$ and $v^T$ is $1\times n$ so the product is $n\times n$. On the other hand, observe that $v^Tu$ is a $1\times n$ multiplied by $n\times 1$ which is a scalar, and it must equal the trace of $A$ (since $\textrm{Tr}(XY)=\textrm{Tr}(YX)$ in general).
Then $$A^5=(uv^T)^5=u(v^Tu)^4v^T=(\textrm{Tr}A)^4 \cdot A.$$
Thus, either $(\textrm{Tr}A)^4=16$ or $\textrm{Tr}A=0$, so the solution set is
$$
\textrm{Tr}A\in \{-2,0,2\}.
$$
A: The eigenvalues of $A$ are roots of $x^5=16x$ and so are in $\{0,\pm 2, \pm 2i\}$.
$A=uv^T$ implies that $A$ has rank at most $1$. Therefore, $0$ is an eigenvalue of $A$ of multiplicity at least $n-1$.
Thus, the trace of $A$, which is the sum of its eigenvalues, is one of the  eigenvalues and so is in $\{0,\pm 2, \pm 2i\}$.
If $A$ is real, then that eigenvalue must be real because there is no room for its conjugate. In this case, the trace of $A$ is in $\{0,\pm 2\}$.
A: We have $A=uv^T$ and so the trace of $A$ is $\tau=\sum_i  u_i v_i = v^T u$.
Then $Au = (uv^T)u = u(v^T u) = \tau u$. Therefore $A^5 u = \tau^5 u$ and so $\tau^5 u = 16\tau u$.
If $u=0$, then $A=0$ and so $\tau=0$.
If $u\ne0$, then $\tau^5 = 16\tau$ and so $\tau \in \{0,\pm 2, \pm 2i\}$.
If $A$ is real, then $\tau \in \{0,\pm 2\}$.
