Finding trace of matrix from given conditions

Question

I have a question that says,

Let A be a matrix such that $A = [a_{ij}]$ such that $a_{ij}=u_iv_j$ $\forall$ $1 \leq i\leq n$ and $1 \leq j\leq n$, and $u_i, v_j \in \mathbb{R}$ satisfies $A^5=16A$, then find $Tr(A)$, $($Here, $Tr(A)$ denotes the trace of $A$$)$

I have tried it as follows:

$$A^5=16A$$

$$\implies A^{-1}A^5=16A^{-1}A$$ $$\implies IA^4=16I$$ $$\implies A^4=16I$$

But, in the question, it is given that $a_{ij}=u_iv_j$, what are $u, v$ here? The answer is stated to be $2$, but I cannot proceed further. Any help on what $u, v$ are and how to proceed would be helpful. I think that the question might have been incomplete, but if anyone can help anyone, it would be nice. I am aware that another query exists regarding the same question, but since that is a year and a half old, and doesn't help me much, I decided to post another query.

Answer 1

The fact that $a_{ij} = u_iv_j$ means that $A$ has rank $\leq 1$ (as $A = U^TV$). Note that this implies that $A$ cannot be inverted unlike what you wrote. Then, we can proceed by looking at the eigenvalues of $A$ (as the trace of $A$ is the sum of these eigenvalues).

The fact that $A$ has rank $0$ or $1$ means that $0$ is an eigenvalue. $A^5 = 16 A$ could be used to find the remaining eigenvalues but we're still missing some information: taking $A=0$ satisfies all your hypotheses but does not yield the answer you expect. This might be solved by having more information on $U$ and $V$ (in particular that they are not $0$).

Answer 2

$a_{ij}=u_iv_j$ is equivalent to saying $A=uv^T$. Hence $$A^2=uv^Tuv^T=(v^T u)A,\quad\ldots\quad A^n=(v^T u)^{n-1} A$$

Given that $A^5=16A$ then shows $v^T u=\pm2$ or $\pm2i$, or $A=0$.
Note that the trace of $A$ is $\sum_iv_iu_i=v^Tu$.

[Note that $A$ is not invertible, so one cannot multiply by $A^{-1}$.]