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> If $n \ge 2$ and if $A$ is a matrix with rank($A$)=1,then the minimal polynomial of $A$ is of degree $2$.
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So, we see that $rank (A) =1$.So, nullity of $A \ge 1$.Now we try to find the range space of $A$.If $y \in R(A)$ then $A(x)=y$.

We see that null space of $A$ is being spanned by $n-1$ linearly independent vectors. We can extend this to a basis of $\mathbb{R}^n$.Let $x\_1,...,x\_n$ be the basis where $x\_1,..,x\_{n-1} \in ker(A)$.We see that $x\_n$ must not be in ker($A$) otherwise $A$ is the zero matrix so that $A=0$.

Let $x\_n$ be the vector which spans the range space Of $A$ such that $A(\sum(c\_ix\_i))=x\_n$ where $1 \le i \le n$.So from here we can conclude that $0,(c\_n)^{-1}$ are the only eigenvalues values. Hence the minimal polynomial becomes $x(x-(c\_n)^{-1})$ .